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0:01
yes I am serious today we'll be doing 100 derivatives and we'll be doing them in one take however this is not the first time doing this though because I actually did this 10 years ago already but let me tell you guys this first I would like to dedicate this video to all my upper bound students especially the ones in 2009 in the summer at UC Berkeley campus and let me tell you guys my upper bound students I still remember everyone I still remember you guys it was such
0:30
a wonderful experience that we had and that was the second time that I had my own class I taught my own Calculus class during the summer and the upper bound you know it's a program that helps high school students who are from low-income family just like myself as well and I of course have to thank my director my main man Mr Mona lumba mum thank you so much for believing me me thank you so much for giving me so many chances to teach the class which tell you of course I
1:06
cannot be who I am today I cannot thank you enough thank you all right so 10 years ago I actually did a PDF because I to teach my class right and I decided to produce you know enough example for my students and how many examples are enough it came out with 100 derivatives questions and also Solutions and them up you guys can find the link to the PDF I will have that in the description for
1:34
you guys but today though we'll be doing something different I have this right here for you guys and notice that the equations here they are in the multiple choice format because I believe it's the best when you have the riv equations with multiple choice format because sometimes you don't know how much more you have to simplify or how much more you have to keep going or when to stop so multiple choice format it's the best for derivative questions and for the people who
2:04
would like to know this yes of course how can we have derivatives without the definition right so if you guys would like the last 10 questions were so yes derivative by using the definition but uh the first 90 question would be just we use the rules that we we we uh have know already okay all that so this is going to be good for K 1 K2 and hopefully you guys all enjoy this and in fact today I have something fun let me just show you guys this right here uh I bought this recently
2:38
right I bought this recently and I opened it this is K the ninja milk m key MK so ninja milk from Ryan hi got and I will be trying his energy drink for the first time so you guys can see I have this package right here right so I have an open the package yet so I will show you guys I will tell you guys how it taste but not now because right now I have enough energy so I don't need it yet
3:12
I don't know how long this will take me though so I don't know how many I'll be drinking so I don't know and of course I have my Marathon metal this is the one from 2007 and in fact PR this meaning personal record uh during this year and you might be wondering why why don't I bring the one that's from 2009 unfortunately I didn't run that one in 2009 all right anyway you guys can find the PDF
3:38
in the description right here all right so yeah all right here we go first question we are going to be differentiating a x s + BX plus C notice that a b c are just constants right so for the first one just start something easy this just the power rule and the idea is that whenever you're adding or subtracting right but this is just all adding you can just differentiate the first and then differentiate the second and differentiate the third and then just put the results together
4:10
that's it so for the first one for this right here I will just bring the two to the front and then minus one that's the Google product power rule right so the first term will get 2 a and then x to the first Power you don't have to put on the one and then the next one is plus when we have B * X to the first Power well you can put on the one right here you can bring the one to the front and then minus one 1 * B is just B and x to the Ser power is just one so it's just B it's just a
4:42
coefficient of x lastly this right here is just a constant so when you differentiate constant C when there's no X the derivative of this is just going to be zero so you don't even have to put that down so for the first one the derivative of a quadratic is just 2 ax x + B and that's it start with something easy right and uh I'm not going to give you guys the answer key hopefully you guys cannot figure this out so uh this a nice pattern right you guys can let me know in the
5:14
comment down below let me know what the pattern is anyway second one we'll be differentiating sin x over 1 + cosine X notice that this is a quo and we'll be just using the quo rule for this so here we go before we forget let's go ahead and square the denominator so we will have 1 + cosine X and we'll raise that to the second power and then the qu rule says we will have to put the denominator
5:45
to the top right here so let me just write down 1 + cosine X and then we multiply by the derivative of the top the derivative of sinx is just going to give us cosine X and of course you have to know all the derivatives of the functions right hm yeah so you have to know that all right so you have all this and then you have to subtract the top function which is sinx and then multiply by the derivative of the bottom function the derivative of one is zero the derivative of cosine X is
6:22
negative sign so we have negative SX right here and now this is the time that you have to think about how we can simplify y all the Expressions right here well of course we can just distribute the top right here so we'll do that do that so cosine * 1 is just cosine x cosine * cosine is cosine squ X and we put the Power two right here right cosine X and then square and then negative negative becomes positive and then sin x * sin x is of course just sin Square X over the
6:57
denominator right here which is 1+ cosine x² and now check this out here we have cosine Square X+ sin sare x guess what yes we can use a nice trick identity for this this right here is nicely equal to one so we just have a one right here in fact on the top we have 1 + cosine X over 1 + cosine X squar therefore what we can do is we can actually cancel the top and perhaps I'll just write it down
7:32
again for you guys to make it clear this is by saying 1 plus cosine X and then we divide it by the bottom is parentheses 1 + cosine X and then to the second power so of course we can cancel this one out with one of the on the bottom and therefore all in all the answer is just one on the top over 1 + cine x to the first power like this and after that we are done we don't need to put down a plus c because this is derivative this is nothing to do with integration right
8:07
so find the answer in the answer choices right there all right number three I will also fit in here right here so we will be differentiating ah it's just the flip of this this right here we have sin x over oops sorry number three I should have written this down earlier number three we have 1 plus cosine X over sinx like this well for this question in fact we don't have to use the quo rule because we only have one thing on the bottom and we have two things on the top
8:41
right two terms we can just split the fraction it will be easier this way so now check this out I'm not doing derivatives yet I will just put down DDX to mean that we will be taking derivative and we split the fractions one over this one over sin x and then the second one is plus cosine X over sin x and then of course 1 / s x do we know anything about that this right here yes it's just cosecant
9:13
X likewise this is just cotangent X so check this out this right here is just to differentiate we haven't done the derivatives yet so we still have to write this down we have to differentiate this is cant X and we will have to add the result with this right here is coent X like this oh my goodness I will have to remember what the derivative of this and that are h ah I think I know the derivative of cosecant X is cose x * Cent X and then the derivative of cang x is
9:53
negative cant X squar or you put a two right here right and that's it you don't have to factor out the cosecant X and you can find the answer in the answer choice right over there so that would be good I I just realized I make a typo on my answer choice and uh I'll fix that later but yeah so yeah if you guys not saw the answer earlier but just uh look at the file in the description I'll fix that before I uh post it so anyway these are the first
10:29
sitations I don't know how long this will take me derivatives usually aren't so bad until we hit the definition right we have to prove the derivative of what and what was the definition of derivatives all right so um three questions down and many more to go and hopefully you guys all find this video to be helpful right all right number four we have the square root right so here we have number four we will have to differentiate the square root
11:09
of 3x + 1 derivative of square root of 3x + one now this is how you can do it first thing first you have to know right you have to know that the derivative of the square root of x like this this right here will happen a lot so you just might as well remember it the derivative of square root of x is 1/ 2 square root of x and to do this you will have to write the square root as the 1/2 power and you bring the power to the front and then minus one from the power so you end up with
11:49
12 x to the - one2 x to 1 half it's the same as putting the square root in the denominator but as I said you should just remember this and then and use that to do the questions like this so here we go to differentiate square root of 3x + 1 we will just have to look at the big picture which is the square root we will have to do 1/ 2 * the square root of the inside which is 3x + 1 right
12:18
it's pretty much the same idea over there however don't forget the chandu we have to multiply by the derivative of the inside the derivative of 3x + 1 is just a three so you multiply by three right here right so this is the chenu and some textbooks they like to call this step the Chen rule so one way or the other they both work pretty nicely in the end of course one * three you can just put the three on the top so 3 over 2 < TK 3x + 1 and guess what we are done so this right here
12:51
is it all right now number five so here we have DDX this again means to take the derivative and we have these two notations sign and you put a three right here of X and then we add sin x like this with a three right here parentheses and technically also have the parth right of there H here is the do this right here right this right here it means we will have sinx right raised to the third power like this and then of course we will have to differentiate this guy right so
13:37
it's just the notation whenever we have the three right here it means we have sin x * sin x * sin x of course we write it like this all right but when we have S of X+ 3 power this means we have S of x * x * X like that parenthesis matters parenthesis matters they are totally different be sure you are familiar with the notations right so this is how I would recommend you guys to do these questions we technically have two questions to do right two parts all right
14:16
here we go we have to differentiate for the first one let's write it as sin x to the thir power and then for the second one I'll just keep it like this just remember to use the CH L uh a few times whenever we need to okay here we go for the first one sin x is a input and then we have this raised to a third power so do the piig picture first bring the power to the front and then minus one so we end up with three parenthesis and the inside States the same for now so you just write
14:51
down sin x and then 3 minus one that's new power which is two and now it's a time to use the can do you look at the inside which is sinx and you ask yourself what's the derivative of sinx which is just cosine X right and you multiply this part by cosine X this is the chain loop and then right here let's go ahead and add for this part well this time though the big picture is sign of this so you differentiate s which is cosine and the input is X power it states the same right here
15:32
okay and then we look into the parenthesis the derivative of X Plus thir power is what well it's 3x squ right just use the power rule so you get 3x squ so this is it right this is it and finally we'll just write this down the usual way so here we go I will put down the three right here and I will put down the cosine first I put that down h no I didn't put on the cosine first I just write down is this right here is the same as saying sign with a two right here right and
16:07
then the cosine X and for this one usually we put the polinomial part in the front so I would write this down first which is the 3x² time cosine of x to the 3 power like this and with that we are done this is going pretty well and like the integrals so this is number number five okay number six I will just put it down right here I'll just erase this a little bit so I can number six here we go
16:42
we have to [Music] differentiate 1/ x to the 4 power like this okay so x to the 4 power this is easy huh so of course we can just look at x to the fourth power bring the four to the front which is just 4X and then my - 1 which is 4 x to the 3 power and that's in the denominator just like that right no don't do it like this no no no no no no this is not the answer not the answer why because the four that you see right here it's not really a four because it's in a denominator
17:22
therefore this right here it's really asking to differentiate X to the negative4 power in order for us to use the power rule remember remember remember we have to have X to some number right well 1 / x four power unfortunately this four is actually a negative4 power so you have to know your rule of exponent really well this is very similar to the square root example you have to write the square root with 1/2 power anyway look at this right here and then do the power rule
18:00
bring this to the front and the minus one right here and we end up with -4 * X to the5 after you subtract them and that's pretty much it finally -5 that's the power for the X so we can bring this down to the denominator therefore we end up with -4 over x to the positive 5 power like this so this right here it's a correct answer this right here it's a common mistake don't do it like
18:32
that okay don't do it like that all right continue continue continue number seven number seven number seven here we are going to differentiate and we have 1 plus cotangent to the third power cotangent X and then to the third power like this right so here we go of course we have the inside and then raise to third power so again we have to do the big picture first which is the outside power rule right bring the three to the front and then minus one so here we go bring the
19:08
three to the front inside States the same just write down 1 plus Cent X and 3 - one is two of course we're not done yet though look at the inside and ask yourself what's the derivative of one which is zero what's the derivative of cotangent X well that's negative cant s x remember the chandu right and of course this right here it looks like subtraction so be sure to just put a parenthesis this is the derivative from here and we have to multiply this with that this right
19:43
here it's pretty much it and in the end we just have to write it down nicely put a negative all the way in the front that's all so finally we have -3 * this which is 1 + Cent X and then square and then in the end I have the Cs C Square X which is the cosecant Square X just like that and that's it all right now number eight I need more space so I will have to erase this now so the first couple
20:17
questions I think the first 20 questions are more for like warm up and I put it like according to the difficulty and also the things that you have to know in the usual order later you will see the inverse trick functions natural log exponential and hyperbolic functions all that stuff all right now things are getting study trickier and also study more complicated anyway here we go number eight we are going to differentiate let's see x² times 2 x^ 3 power + 1 and raised to the 10
21:02
power like that all right this one compared to the earlier one well we have X squ right here times this quantity to the 10 power so it's not just a simple chain rule question because right here we will actually have to use the product rule because we have X squ time this and this is how I like to set up the power this is how like to set up the product rule for my students for the first I'll show you guys at this one time right here in this video so here we go well this right here
21:37
it's the first function so I'll just write down X squ and I will call that to b f and this is just the first function X squ like this and this is a second function so it's G which is 2x^ 3 power + 1 to the 10 power like that well separate this and I will differentiate them individually for the the first one the derivative of this which is just 2X and then the derivative of this this is G
22:03
right so G Prime is going to be the following the derivative this of course you bring the 10 to the front minus one and we get 10 and the inside States the same 2x+ power + 1 to the N power like that and then we multiply by the derivative of the inside because of the Chad L the derivative of the inside it's just pretty much this part we need the 6X right here because the derivative one just zero so it's just like that and of course we can simplify this a little bit 6X * 10 so 6tx
22:39
and this is 2x^ 3 power + 1 to the 9th power so you can just set the like this and then we just kind of cross multiply to construct our answer product rule in action f * G Prime so we will have x² time this right so let me just put this down right here which is 60x * 2x^ 3 power + 1 to the 9th power and then we go the other way which is this times that we add this times that which is the original in blue we have parentheses 2x+ power + 1 to the 10 power and we multiply by 2X
23:30
so this is the write out right by using the product rule and also the chain rule so you just have to do it carefully and later I'll show you guys without these things right but I think this is helpful especially if you're doing derivative if you're taking calculus for the first time all right in the end we just have to simplify this let's see first of all we notice that both terms they have this Factor right 2x to 3 power + 1 and this is n9th power and this
24:03
is the 10 power so we can factor out the smaller exponent so perhaps let's do that and in fact we can also Factor things of this right here is 2X and this is 60x both of them have X and both of them have a factor too so perhaps let's do this first let's put down the factor of this right so let's factor out 2x right let's fact 2x okay in the meantime we can also Factor this out which is 2x^ 3 power + 1 to the 9's power like this and now let me show you what the left over is
24:42
well if you look at this part we Factor the out of 2x so 60 / 2 is 30 this x is out but we still have X squ so we write down X squ right here and you see this part we Factor the already so this is all we have we have Nam the 30X s now if we look at this part first we have to add notice the 2x is all the in the front already and then for this part we see we took out nine of the factors already that means we have one more so we just have the inside which
25:15
is 2x to the 3 power + 1 to the first Power just like that and in the end I just kind of
25:32
leave it like that I think yeah I didn't do any ah no no I make a small mistake man okay okay okay okay 2x+ thir power the D of 2x thir power is 6 x² my bad right and then this right here should have been a 6 TX square right here 6tx squ right here and uh let's see so this right here should have been 6tx s and let me just do this part again for you guys let me just do this part again for you guys all right so 60x squ right here again same thing we can factor out the 2x
26:19
2x right so we can factor out 2x let's do that and we can also Factor this out which is 2x + 3 power + 1 to the 9th power and now actually I put this down in blue earlier so I'll put it down in blue again anyway 2x 2x to the 3 power + one and then to the 9 power and parenthesis you see I make mistake too okay parentheses okay for this part again we took out the two so we have 30 and then we took out X so this right here is just x to the first power x * x s though is X to thir
26:56
power again because this right here it's X squ so it's x * X so you took out one of them you have another X this x * X to second power we get X to 3 power all right and then that's it for the first part because this is out for this part this is out we have this left to the first power so we add 2x^ the 3 power + 1 to the first Power like this in the end we can combine this and that together so we'll do that and finally we see the answer is where's my equal sign there we go
27:33
the answer is 2x * 2x to 3 power + 1 to the 9th power times this plus that which is 32 x to the 3 power and in the end we have the plus one and that's it so it simplified pretty nicely right and yeah all right now number nine and by the way took me a lot of time to come up with the answer choices as well not just the question I want to make it rather legitimate I can just like give it to my students for them to practice and for the actual exams or actual C that I give out to my
28:17
students I usually have about five or six answer choices but this is just for practice that's why I only have three choices anyway number nine here we go we have this we have to differentiate X over and the bottom here is parentheses X2 + 1 to the second power like that well well why don't I show you guys this right here as well I call this the memory device whenever we have to do the product rule equ rule and chain Rule and all that especially if the question is complicated I
28:55
like to show my students this especially for the first time anyway here we go this right here is a quotient X over this right so we have to use the quo Rule and the quo is just the opposite of the product so what you do is on the top is the F function on the bottom is the G function well earlier you see the first one is f the second one is G well for the quo rule we actually write down
29:20
G first and then F right here so the G function is this which is x² + 1 to the second power and then the F function is just X on the top once we put them down individually we can just differentiate them individually for the first one we will just get G Prime and this is going to be okay put the two to the front minus one and we will get two parenthesis X2 + 1 to the first Power and we multiply by the derivative of the inside because of the chandu the derivative of this is
29:55
2x so just like that well for this one of course differentiate this real quick we just get one so that's pretty nice and let's simplify this a little bit this is just 4 x * x² + 1 if you would like right so I think we'll just leave it like that now the answer to this to construct the qu R you do this time still right but before we forget we have to square the denom right away so I will put down parenthesis with this inside x² + 1 s this had a square that's the original but
30:38
we still have to square it because of the quot rule right okay for the top this is where the top comes in I'm going to multiply this with that so this right here is just that which is I'll put in I'll just put in real why not X2 + 1 to the second we just put in Black this times that's x sare + 1 s and we multiply by one right here and then for the quo R remember it's subtraction so we subtract and we go from here to here so we subtract this times that all right we have
31:16
x times this part which is parentheses let me just put on dot * 4X * X2 + 1 like this and of course this is the good way you can just Square the denominator and you put the button function right here and you multiply by the derivative of the top and you minus the top function times the derivative of the bottom and later on I will really just be doing that steps without showing this right here okay let's see if this anything we can do to simplify this notice on the top we
31:55
can again Factor things out because here we have X2 + 1 to the 2 power and here we have X2 + 1 to the 1 power so we can factor that out parentheses x² + 1 and let's see what we have left originally this was to a second power we took out one of them so we still have x² + 1 good and then we have to minus well this is out but we still have to have this and that and we multiply them together 4 x squ right here so we put that down of course and then all over of course when
32:36
we have this square and the square we multiply the powers so we end up with x² + 1 to the four power and now check this out this and that can reduce so we'll cancel this and that out and this becomes the third power and that's pretty much all we can do and of course we also combine this and that so finally this minus that is -3x s and then we have the + 1 over this to the thir power so x² + 1 to the thir power like this and with that we are done that is it and very nice
33:20
we did the first page already let me just show you that was the first page but the problem is yeah 13 pages so I don't know man anyway uh 10 questions took me this all five six hours again it's okay though the more we do it the faster we get anyway number 10 right but uh I will have to erase this okay
34:00
maybe I will drink the ninja milk milk the ninja milk um once I reach equation number 30 yeah I think that's a fair uh thing for me to do H all right number 10 differentiating the following so we will be differentiating on the top we have 20 and on the bottom we have 1 + 5 and this is e to the -2X like this right like that all right good well let me just take us little thing this right here like
34:44
this function right here you can see it um from the logistic growth function so it's not quite exponential but it's logistic so it's really legit okay well how do we do differentiate that you don't necessarily have to use the qu rule you can but you don't have to and the reason is because notice on the top we just have 20 which is just a constant this is the function part right here right it's easier to do it this way we can just write it as to take the derivative 20 is still 20
35:19
but we can write this the denominator bring that up and raise that to the negative one power so 20 * 1 + 5 e -2X like this to the -1 power and the reason to do that is because we don't have to use quot rule this way we can just bring the negative one to the front and then minus one to the power and do the Chand do after that all right 20 * 1 we have -20 and this right here again States the same
35:56
1 + 5 e -2X 1 1 Al together you get -2 but not done yet use the CH do multiply by the derivative of the inside derivative of 1 is zero derivative of 5 e to the -2X is the following whenever we have e to this part just repeat itself first so we multiply by 5 e to the -2X but we're not done yet because we have to look at the derivative of the inside of the inside which is this part the
36:31
derivative of -2X is -2 so we have to multiply by -2 right here so we have to use the chain R twice for this part right so this is from here and then to differentiate this part we have to multiply by the derivative of -2X which is the2 right here okay now we will just put everything together let's see negative negative we get positive 2 * 5 is 10 * 2 uh * 20 is 200 and I will just leave
37:08
this on the top as well so we have 20 and I mean 200 and we have e to the -2X like this and then I have this all over we can bring this down to the denominator which is 1 + 5 e^ -2X this negative 2 is the second power like that and that's the answer choice right here and I just stop it right here sometimes maybe we'll write this down here but it depends I think this is more clear on
37:42
using this method or maybe using the question rule but as I said whenever you have a multiple choice question of course we choose the best answer so that would be it now number 11 we have to differentiate here we have e to the X in the square root and then for the second one we have to add e to the square root of x we are not using the product rule right here this is just addition so we differentiate the first and then differentiate the
38:16
second well check this out whenever we have square roots in calculus we almost always look at square roots as one half power so in fact for the first one we can look at this and we're just doing algebra so the idea is that we should be doing algebra first before we do calculus so if you look at this part we just look at this to the 1/2 power and this is e to the X inside this part States how it is and the reason I want to point this out is because this right here
38:53
we can just do x * 12 and we can re write this as to take the derivative of e to the 12x plus e to the square root of x like that do Algebra first before you do calculus and then right here we can just do the calculus okay here we go let me actually just probably put it here like that this is not integration side this is just an arrow right anyway here we go differentiating e to the something we first just repeat the whole thing namely e to the 12 x and then we do the Chand do
39:37
look at the exponent right here and ask ourself what's the derivative of 12 x and the answer to that is just2 good now we do the second part and notice again we're just adding so we just have to add the result for the second we are not doing the product rule right the derivative have e to the root of x we can just first repeat that and then again I look at this right here and ask myself
40:04
what's the derivative of square root of x we talk about this earlier so that's why I didn't you know write that as the 1/ half power derivative of square root of x is just 1 over 2 root X like that and that's pretty much it in the end perhaps we'll just write this down nicely check this out e to the 12x this part is this which is that which is that so this part is just the square root of
40:34
e to the X inside over two so I'll put down over two right here and then for this part I can just put this on the top this times one so we put on plus e raised to the square root of x power / 2 square root of x like this of course you might want to get the common denominator if you want and all that but again I left the answer like this in the answer choice so that's it for the question so that's yeah okay now number 12 what do I need to do oh this is a good one number 12
41:19
differentiating here we are going to differentiate secant to the thir power of 2 X secant to the 3 power of 2x well it's a good idea to look at this as to take the derivative of parenthesis secant of 2x inside like this and then to the third power especially if this is the first time that we're doing derivatives it's always a good idea to put the power right here and don't forget to use the big parenthesis to group The Expressions all right here we go we have to look at this to
41:59
the thir power so we can just put down the power to the front and then minus one so the first part we have three and then we have this right here which is parentheses secant of 2x the inside stats the same for now right stays the same right here and 3 - 1 is two that's a new power and we do the chandu look at the inside this is the time that we have to do the derivative the inside derivative
42:28
secant is secant of the inside time tangent of the inside right so we will have secant of 2x again the inside States the same and we also have to have the tangent of 2x like this but what's the inside of the inside it was a 2X isn't it well we have to do the chendu again what's the derivative of 2x just a two very nice multiply by two now we'll just put everything together 3 * 2 is of course six so Al Al together we have six and check this out this is secant to the second power
43:11
and secant to the first Power they have the same input right this times that together we can write it as secant to the third power and the input stays the same and we multiply that with tangent of 2x just like this so this right here is the answer second power first power together we have the third power and we put the power right there for the final answer just like that okay number 13 it's one of my favorite ones right you you'll see why so number 13 number 13 number 13 we will
43:49
be differentiating 12 secant x tangent X Plus and this is the first time that we have the natural log so I'll talk about the natural log function 12 natural log of secant x plus tangent X like this so parenthesis and another parentheses for that all right here we go there's nothing that we can do to rewrite unfortunately yeah this is inside of the natural law so we have to just deal with
44:25
these two parts individually for the first part we have to use the product rule because we have the first function times the second function so let's get this part done first well product rule in action we have to keep the first function which is one half and we have see X this is just the first function times the derivative of the second the derivative of second the RO of the tangent X is secant s x and again you have to know your derivative 3D well for all this kind
44:59
of questions and later on in C two or maybe when you're doing Integrations in C one you have to know the derivatives backwards anyway first one * thetive second plus the second function which is tangent times the derivative of the first and notice I have the 1 half see X as my first function so I need to have the 1/2 whenever we have a constant multiple just go ahead and put it down and then you focus on differentiating the secant x which will give us secant x time tangent
45:38
X so this right here is the first part okay now I will do the second part right here in blue we will have to add well again 1/2 times this thing so we will have to just maintain the 1/2 as 1/2 and this right here as a reminder whenever we differentiate natural log of x natural log the inside we just get one over the inside like this and of course we will have to use the chain rule right okay
46:10
so we have the natural log of this so we do one over this thing which is secant x plus tangent X well well CH do right here isn't it we have to multiply by the derivative of the inside right okay this right here it's a plus so you just do the derivative of the first and you add the derivative second you have two things so be sure you put the parenthesis the derivative of the first is secant x plus time tangent X der secant x is this and then you just need to add
46:46
the derivative of secant x sorry the derivative of tangent X is secant squ x like this wow such a long WR expression isn't it so again the derivative of secant x is that we add it with the Der tent X is that so this is what we have but do not leave this as the answer because we can do better than this check this out to see what we can do hm uh this is kind of weird I'll show you check this out here we have secant secant Square so we can factor out secant isn't it so
47:24
we can factor out secant x and then here we will have tangent X Plus secant x right if we Factor the out hey look at that here we have secant x plus tangent X here we have tangent X Plus secant x and of course the order of addition doesn't matter therefore we can legitimately cancel this and that out so for the second part we actually just end up with 12 secant x for the derivative that's very nice and perhaps let's look at this first part as well hm we have okay
48:10
for this part we will just write this down in blue okay I will just put this down right here we add 1 12 secant x again that's just the first that's just the second part and now we still have to take care of this part part you just have to we still have to take care of this part and it seems like we can also Factor things out huh because this right here has 1/ 12 x well 1 12 secant x and this is also 1 12 secant x so let's Factor it out and see what happens 12 secant x right
48:48
in fact we can also Factor this out but let's do that later because it's kind of slighty too confusing a little bit left over here we have secant Square X Plus left over tangent tangent so we have tangent Square X that's it okay so that's pretty much the factorings that we have at the moment and now of course we can Factor this out together so we see this and that can be factored it put it outside right put it outside take the secant outside multiply with
49:25
a left over here we have secant Square X and this is plus tangent Square X and we have to add well we Factor this out in the front already so for this we have a one left so that's what we have but now is there anything else that we can do in fact this is not one of the answer in the answer choices H this is the time now we have to know our trick identities unfortunately secant Square X Plus tangent Square X we cannot combine them nicely but we can actually combine tangent Square
50:08
x + 1 nicely because this right here gives us nice D SEC Square X as well I'm serious this add up to be secant Square X so check this out all in all we have 1 12 secant squ we have one have secant x in the front right and for the inside we have secant squ plus another secant Square so we have two secant squ x like that again this plus that finally you see 1 12 * 2 well of course they cancel and lastly secant * secant to the second power is of course secant to the third power so
50:50
congratulations this right here is simplifies to secant to the thir thir power x like this yeah see if you differentiate see to third power you get that for the answer like 2x inside if you differentiate this you get C to the thir power x for the answer and let me tell you in C2 you actually have to know how to go backwards meaning you start with secant to the thir power you have to figure this out right here so that's it and then as you guys saw I Dro my marker so hold on
51:27
I got too excited but anyway this right here is it secant to the 3 power x right okay that was question number 13 so hopefully you guys can see some of the questions are going to be slly trickier than the others but some of the questions is just that if you know all the details you should be able to put the answers together nicely quickly simply um yeah and all that stuff all right now was question number 13 okay here we go number
52:10
14 so differentiating on the top we have x * e to the X and notice as this on the top because we have 1 plus e to the X on the bottom oh my God like that anyway we can do it don't worry now this question requires both the quo Rule and also the product rule because this is a quo right this over that and on the top x * e to the X it's a product so we have to do both anyway check this out first of all before we forget let's just Square the denominator and
52:54
I will just give myself a little bit more space let me just put the parenthesis down right here and 1 + e to the X inside and we square that now the quo rule says I will have to put the botton function right here which is 1 + e to the X and we multiply by the derivative of the top so here we go open the parenthesis because we have to use the product rule and we have to two terms right so product rule in action for the top we keep the first function which is X and we
53:31
multiply by the Der to the second derivative e to X is just e to X so we have that and we add the second function which is e to x times the derivative of the first derivative X is just one so we multiply by one like that but we're not done yet minus okay I need this to be longer put this St more in the middle 1 + e to X okay anyway minus we write down the top function which is x e to the X and then we multiply by the derivative of the bottom derivative of one is zero derivative e
54:10
to the x is just e to the X so we multiply by e to the X like that so this right here it's a write out it's not so bad and we just have to simplify this um yeah okay okay let's see what we have right here so just just just kind of don't pain anything just just just do it anyway we do one times that I will put down the result on the top it might be easier this way so one times this is just x e to the X and the one times that is just e to the X right next e to the x times this
54:49
we add X and e to the x * e to the X we add the exponents so we'll get e to the 2x right so e to the 2x lastly this times that so again plus e to 2x good then we minus this times that well we have the X right here e to x * e to X is just e to the 2x so that's pretty nice well well is there anything that we can do to simplify yes because this right here and that they are exactly the same by opposite sign so we can cancel them out and uh that's it so that's pretty much on the top so
55:36
in the end we have x e to the X Plus e to X Plus e to the 2x we can Factor things out but um the answer is just in this form so we'll leave it on the bottom just that 1 + e to 2X sorry 1 + e to the X raised to the 2 power just the bottom and with that we are done all right number 15 D DX we will have to differentiate e to the 4X time cosine of x / two right so notice that this is a multiplication so be sure we use the product rule and this is number 15 number 15 guess number 15
56:38
okay number 15 number 15 is not number five there we go okay here we go this is just going to be a product rule example well we of course keep the first function which is e to 4X and we multiply by the derivative the second well we have to use the chain rule right here as well so don't just be careful anyway the derivative of the second function derivative of cosine is negative s so we multiply by parentheses negative s and the input States the same so x/ two right here but
57:15
we will have to multiply by the derivative of the inside what's the derivative of x/ 2 it's just 12 good why this is just a small note okay because I have experience some students they would like to just use the qu rule right here you don't do the qu rule for that because when you have to differentiate X over two you have two ways to do it you can just look at this as to differentiate 1
57:44
12 x right 1/2 * X and of course you just end out with 1/2 like that so differentiating this is just 1/2 do not do quot rule whenever you have a number on the bottom just focus on the number one another thing you can do is bring the fraction one2 in the front and then differentiate whatever function is left just like that and you see this is 1/2 time the derivative of x is just one so it's just 1
58:16
12 * 1 which is just 1/2 I just all that stuff yeah so yeah all right uh I have to close the parenthesis right here not done yet nothing yet this is just the first function times the Der to the second and then we add a second function so we put on plus and the second function is cosine of x / 2 times the derivative of the first function so multiply by e to the 4X right because the Der of first function the the E to 4X we first write it down and then we multiply by the derivative
58:57
of 4X which is just a four so this right here is the chu part so this right here it's pretty good and there's nothing too much we can do to simplify and I didn't factor out e2d whatever so I will just pretty much write down the results that's how it is so finally the answer right here is2 e to the 4X Time s of X /2 and then 4 all the way in the front so plus 4 and then let me put down e to 4X and cosine of x / 2 like that so this right here is the answer for
59:46
that derivative all right number 16 I will just write it down here number 16 here we are going to differentiate 1 over the fourth root of x^ 3 power minus 2 like that so we use the qu rule no because on the top again it's just a number it's just a one right so it's a much better idea to write this as a negative power and this right here it's like the same say 1 over 4 power so here we go this is the fourth root it's like one over four Power by it's in
1:00:34
the denominator so when you write it on the top we will just write as parentheses with X to the 3 power minus 2 inside and then we raise this to the - 1 over 4 power like that and when we are looking at the expression this way we can just use the power Rule and also the chandu and we will do that of course so here we go uh I'll just put this down blue why not all right power ruing action first so
1:01:08
we have thetive 1 over4 all the the front inside States the same so x + power minus 2 - 1 over4 minus one just do your fraction you end up with -5 over 4 and we do our chandle the derivative of the inside the derivative of x^ thir power is 3x s and just one term because the derivative of -2 is just zero so this is all we need lastly we ping St together negative and this is on the top so let me write down as -3x s over we have this four right here I'm going to bring this down
1:01:52
to the denominator and because that's that's a negative power right so we'll just write as x^ 3 power minus 2times what two and then raised to the 5 over Four's power and that's a positive 5 over four well I didn't leave the answer like that I changed the answer to the radical form so remember whenever we have a force like if the power has over four that turns out to be a fourth root so
1:02:22
on the top is -3x s over this is four this little fractional denominator is the fourth root and then we just write inside as X+ 3 power minus 2 and raised to the 5th power and some people might be wondering should I factor out should I take out X power minus 2 to the first Power and then times doesn't matter this is the answer already okay you can simplify the radical if necessary by in calculus this right here it's totally acceptable all right number 17 I will do it here
1:03:00
maybe uh let's see let me just write it down right here I don't know why I want to do this but like I think I can fit in number 17 right here before I erase so let me just do that I will erase this part and then I'll put the number 17 right here I know I just feel I I I just feel really satisfying if I can fit a lot of math on one board no anyway we are going to be differentiating this is the first time that we're doing the inverse tangent so we have the inverse
1:03:39
tangent and this notation I know it's one of the most disliked notation in math some people like to write it as the arc tangent but it's the same thing right so just get used to it this doesn't mean one over tangent so just keep that in mind anyway we have the inverse tangent and the input is square root of x² - 1 like that and of course if you write as the r tangent you just write R tangent of square root of x square minus one like that okay anyway first of all this right here is
1:04:14
not one over tangent of whatever you don't do the power rule for this you just have to know the derivative for inverse tangent so I put down a quick note right here for you guys today if we differentiate inverse tangent of X this is the one that you definitely have to know really well you get 1/ 1 + x² right well how do we prove it you'll see by the end of the video numbers I don't remember anyway here we go so whenever we're differentiating INF tangent of the input
1:04:51
you see that we will do one over 1 plus the the input being squared so we'll do that right here we do one over 1+ we put the inside right here and square that so I will put down parentheses with square otk of x² minus one inside and I'll square that right so that's the general form for the inverse tangent derivative but again we will have to do the chu here we go the derivative of square root of something is 1 over two square root of the inside you keep it again right just keep
1:05:29
it but we do the CH do again because the derivative of the inside derivative x squ minus 1 is 2x so we multiply by 2x like this all right so this is the idea this is pretty much it and now we just simplify I noticed that we have the two over two so we can cancel them out that's very nice and we also see that this square and the square root cancel each other out that's also very nice when we cancel this out we get one plus this is just x² minus 1 1 - one
1:06:05
cancel the out that's pretty nice so that's what we have is there anything else that we cancel out yes because we see that this is just X over X squ right so let me just write it down nicely for you X over this right here is X squ and we keep it Square < TK of x² minus 1 this and that we can cancel them out we'll just get one and this becomes the first power so finally the answer is one over only x to the first Power in the front and then square otk of x
1:06:43
s minus one that's the answer right here okay cool now number 18 number 18 number 18 number 18 and number 18 this is going to go on for six hours for sure hopefully at some point I can do like a I can speed up the process but as I said the first 20 questions was so I just want to make sure um I give that detail explanation all that anyway number 18 here we have to differentiate we have natural log of x over x to the 3 power like that all right here
1:07:32
we go X to thir power in the denominator and a natural log of x on the top we have to use the quo rule two functions dividing quo rule some people will tell you to write this as X to3 Power Times natural log of x and do the product rule that's fine up to you though I'll just do the question rule when not so let's go ahead and square the denominator so we have X to thir power and we square that on the top we keep the bottom function which is X plus power
1:08:06
and we multiply by the derivative of the top the derivative of natural log of x is 1 /x good and then we minus the top function which is natural log of x I would like to use a parenthesis right here natural log with the input X because I will have to multiply by the derivative of the inside the derivative of sorry the derivative of the bottom the derivative of x to3 power is 3x² without the parenthesis it seems like you can just multiply the X and 3X squ together but no
1:08:44
this x is an input for the natural log this is just like 3x s be multipli with the whole function you cannot multiply the X and the X squ right here no right anyway this is pretty much it for the derivative now simplify this X+ 3 power over X we get x to the second power let me just cancel things out for you guys this right here cancel out so we get x square and this is minus and perhaps I'll just bring that to the front 3x² and then we have natural log of x whenever
1:09:19
there's nothing after the natural log and there's only one input for the X for the natural log you don't necessarily have to put down the parenthesis so this is okay on the bottom X Power Square we get X+ 6 that's the answer you'll see X+ 6 well we can do more because notice that we can factor out the X squ and then cancel things out right so we will do that we have the x square in the front and then 1 - 3 natural log of x over x to the 6 power we can reduce that's how I know this
1:09:58
a for anyway finally we get 1 - 3 natural log of x over x to the 4th power and we are done that is it okay number 19 we are going to be differentiating x to the X power
1:10:27
ah I see X to X power have the power FR I can just bring the power to the front and then this is X and then one minus one for the power x - one like this of course not you know it you know it you know it otherwise I have to change that to Blue huh all right here is the deal this right here it's not a power function and it's also not an exponential function it's kind of like both because you have X for the base and you also have
1:11:02
X for the exponent right so it's like both I will show you guys how you can take off this without doing the implicit differentiation this is the deal the idea is that you want to work with space e right whenever you have exponential function but this is not really exponential function yet because again the base is X you don't want to work with Bas X this is how you can do it do X and I'll put on a note right here for you guys what we can do is we can write X as e as I said
1:11:37
because we like to work with base e rather than any other base and well if I want to get back to X I can just put e to the natural log of x power because they cancel you just get X back now we'll just change this we will be looking looking at e to the natural log of x and then raised to the X power right well however you see that this to X power we can just multiply the powers together
1:12:08
so we just have to differentiate e for the base x times natural log of x for the power so this is the idea you want to work with bace E when ever you have a function to a function power but this is one of the ways to do it I will show you guys the implicit differentiation once we get to that part anyway here we go to differentiate e to the something first of all just repeat that which is
1:12:40
just e to the same thing X natural log of x and then you multiply by the derivative of the input because of the chu right derivative this I will keep the X keep the first function and we multiply by the derivative of the second derivative of natural log of x is just one /x and then we add the second function which is natural log X times the derivative of the first derivative of x is just one so multiply by one like that so that's pretty much the idea and now check this out what's
1:13:19
e to the X to the natural log of x power this is the same as that yeah which is the same as that which is the same as X to X so all the way in the front we have x to the X power and then x * 1 /x of course is just one and this right here is just natural log of x and we're done that's it x to the x * 1 + natural log of x that it's the answer pretty good huh and now as I said we are getting to the implicit differentiation part for question number 20 to 29 I believe that's my set
1:13:59
up for that and I know my my hands are really dirty because of all this is pretty dirty so yeah uh let's see I shall be able to do here all right question number 20 okay we have to find D d y DX right so our goal is to find Dy DX which is the derivative but the problem is that the Y not by itself we will have the equation x^ the 3 power + y the 3 power and we have that equals we have that equal to 6 * x * y right so we are given this equation the
1:14:48
Y is not by itself but it's okay this is how we can do it let me just D let me just put down DDX all the way in the front to tell people that we are differentiating with respect to X right and with that being said that means Y is a function of X and you have to do the following carefully anyway here we go differentiating X to thir power we get 3x² and that's it differentiating y to the 3 power we get + 3 y square but you know this is like saying y to the thir power so we have
1:15:29
this and then you look at inside the derivative of y to the 3 power was 3 y s but since Y is a function of X we will have to multiply by Dy DX the derivative the inside derivative Y is dy DX and you might be wondering how come we didn't do that well if you were like you can just think about like this X Plus thir power so you have 3x s and you should multiply by DX DX huh isn't it well what's DX over DX just one so you don't have to do that so be really
1:16:05
careful with this idea so this is just that but this will give you this right here okay now for the right hand side this is 6X * y so let me just say this is the first function that's the second product rule in action keep the first function which is 6X and multiply by the derivative of the second the derivative of y is just Dy DX and then we add the second function which is y and we multiply by the derivative of 6X which is just a six my God this right here it's a write out
1:16:43
for the implicit differentiation and now we have to isolate Dy DX because that is the derivative and to do so well I'm going to move this to the other side I'm going to move this to the other side right because I want to collect all the terms that have the dydx so we can factor out the dydx and then can isolate that well let me just write this down for you guys I'm going to move this to
1:17:12
the other side but I will also Factor the dydx in the meantime and when I factored it out I would like to put the dydx at the end and then open the parenthesis in the front and the reason I want to put the dydx at the end is that this right here it's the object it's the derivative it looks similar to DDX but DDX means to take the derivative right this is what we are trying to find already this is the things that you have to do the operator so I prefer to have this at the
1:17:47
end of the parenthesis rather than in the front because otherwise it might look like we still have to differ which is not correct anyway this right here was positive 3 y s I brought this to the other side so you be negative with a 6X and we factor out the dydx at the end right there already this is equal to 6 y let's keep the positive first and then we bring this to the other side becomes -3x s like that so that's pretty much it huh and then of course we can divide
1:18:24
so we see that Dy DX equals here we have 6 y - 3x 2 and over right here we have 3 y^ 2 - 6X well we can actually reduce this a little bit and um I'm just checking that my answer key is the jet yes it's good my answer key is good because as I said I come up with all the answer key no I I come up with all the um answer choices myself I just want to make sure I didn't mistype anything so anyway
1:19:00
we can Factor our three on the top so here we have 2 y - X2 and we can Factor our three on the bottom as well so this is Y 2 - 2x cancel cancel and in the end Dy DX equal 2 y - x² over y^ 2 - 2x I just notice if you differentiate this you get that if you differentiate that you get this I don't know coincidence yeah that that's that's it oh interesting enough I actually have this right here right number 20 you guys
1:19:48
see it right here I have the ENT key backwards um on the answer key the reason I did it is I just put the X first and then Factor it out so in fact this right here it's the same as saying we can Factor negative and then put the x s minus 2 Y and then over negative and we put this first which is 2x minus y^ 2 and cancel the negative and it depends how you move the things around so it's either this one or this one and you should be able to see that
1:20:25
the answers are equivalent so that's just algebra part that you have to know so this and that are the same right so yeah just like this so take a look take a look take a look all right I um was expecting to and I think to do 100 derivatives will take longer to do compared to 100 integrals because uh it's just like so many things I have to go through I I don't know man I don't know which one take me longer I don't know anyway next
1:21:13
one another implicit differentiation I will just write down the equation for you guys though so here we have number 21 should I retire this marker I hope not
1:21:25
good job you can still work okay we have y sin y equal x sin x all right okay y it's not isolated and in fact you cannot isolate y in this situation it's okay just go ahead do DDX implicit differentiation here we go this right here requires the product rule so I will keep the first function times the Der to the second derivative of s of something is cosine of the input and then the input is y so we have to multiply by dydx because that's the derivative of y and this is the chu and
1:22:13
then we have to add the second function which is s y right second function is s y so just put on S of Y and we multiply by the derivative of the first the derivative of y is dy DX so we multiply by Dy DX like that so this is the left hand side now for the right hand side same thing but with X instead of Y product rule again keep the first function times the Der to the second theity of s
1:22:48
is just cosine and notice that the input X and we are differentiate that with respect to X so it's just cosine X like this and we're done for that okay and then we add the second function which is sin x and we multiply by the Dera to the first which is one so I should also put down the cos x first times the Der second plus the second times the derivative of the first this is it and now we
1:23:17
can just factor out the dydx on the left hand side and then divide it that's pretty much it right so we'll do it in one step because we see that this is just Dy DX we can just have this and that being factored it in the parenthesis this is the right hand side so we have x cosine X Plus sin x over this and that in the denominator namely y cosine y + sin Y and we are done this right here is the answer and that was number 21 and this is number
1:24:02
21 all right number 22 here we are going to differentiate natural log of x over Y and this is equal to e to the X Y to the 3 power like that okay so implicit differentiation but before we do anything like that notice that we have a natural log of x over y the truth is we can break this down we can write this as natural log of x minus natural log of Y so that we can just differentiate the first and then minus the the root to the second
1:24:44
right and of course this is just the algebra we'll just write down the rest and here we will just put down DDX to simplify to signify that we are differentiating with respect to X here we go derivative of natural log of x is just 1 /x that's it minus the derivative of natural log of Y is 1 over y but the inside is y the derivative y with respect to X is dy DX this is how you squeeze out the Dy DX equal here we have the Der of this well
1:25:26
e to the you just repeat e to that whatever X Y to the 3 power and now we differentiate the power yes we do the product rule this is x * that right so the first function is X and we multiply by the derivative the second derivative y to Ser power we get 3 Y 2 and you know the do we have to have the Dy DX and then we add the second function which is y + thir Power Times the derivative of d x which is just one so this right here is what we
1:26:05
have well well let's see what we can do with all that this is not this is man we have to simplify this well we have a fraction with X and Y right X here and Y here for the denominators I think it's a good idea to just multiply out everything by XY so that we don't have to deal with [Music] the complex fractions later on so we'll do that and perhaps I'll just put down this in blue okay so
1:26:43
we are going to multiply Everybody by XY so again the point is so that we don't need to worry about the common we don't need to worry about the complex fraction St on all right XY * 1 /x we get Y and then minus XY * 1 / y would get X and this is still dydx and perhaps I'll just put on XY right here so we have XY e to the XY to third power and uh then let's also multiply everything else right here so check this out okay let's just do this backwards right so it's a this times that
1:27:33
we did this times this already oh just man draw the arrows is kind of fun we did this part already now we have to do this times that but also this right here has to be multiplied inside okay so here we go x * X is x² y * 3 Y 2 so we have the three right here and then y^ 2are * Y is y Cub right yes and then I will put this down which is e to the X Y cub and I will put down the dydx good next we do this times that so we put down plus we have X Y to the four power
1:28:17
and multiply by that so it's e to the X Y cub just like that very good huh all right now we have to do what we did before I'm going to keep this on the right hand side because I would like to add this to the other side I want to keep everything positive right and now bring this to the other side it's easier this way so when we do that let's see on the right hand side again we will Factor this out we'll put it at the end so we have the Dy DX well this this was positive so
1:28:53
we just write down 3x² y 3 power e to the X Y to 3 power and when we bring this to the other side it becomes a plus X and this is being factored out right here already so that's what we have and for the left hand side this Y is still just chilling right here but we will have to move this to the other side so we minus X Y to the 4th power and then e to the X Y Cub like that in the end we just need to divide this on both sides and we'll be done so perhaps I will I should be able
1:29:32
to fit in here so dydx equals this right here will be divided here so this is on the top so I will just write it right here y minus Yus X Y to the 4th power e to the X Y to the 3 power and then divided by well um perhaps we'll put this at the end just put the X right here so we have X plus this part 3 x² y Cub e to the X Y Cub like that and then me just make sure I didn't do anything wrong I didn't do anything wrong because I see the answer my
1:30:22
key good so a slightly longer one but it's not that bad so take a look right here all right question number 23 question number 23 this right here is a weird one we have x equals secant of Y like that and now you might be wondering okay can we just isolate the Y right away by taking the inverse secant on both sides yes you can do that but what's the derivative inverse secant well if you remember it's okay you can just put on the answer well h i I'll just tell
1:31:08
you guys that I do know the answer I'll just put down right here uh derivative of inverse secant of X this right here is 1/x otk of X2 - 1 and does this look for to you guys no I'm not trying to show you guys my back I am trying to show you guys my back but this is actually one of the answers that we got earlier from one of the questions right you can let me know which question was anyway if you remember this good but if you don't this is how you can prove it I will
1:31:40
show you right here the one's not isolated this is slly more preferred it because we do know the derivative secant x see Y in this case actually so we can handle the original secant this is how Okay taking a derivative X is just one well derivative of secant this is the original we just have secant of the input times tangent of the input and the derivative of the input thanks to the chandu we have to multiply by Dy DX and now we can just keep this right
1:32:19
here and divide both sides by this so we get dydx being equal to 1 over SEC y times tangent y right and you might be thinking this is not that bad at all but here's the trouble if you want to figure out the derivative of infers secant x well if you want to just take the infers secant x on both sides the idea is that when you take the derivative of inverse secant of x your answer should be in terms of X as well isn't it so now this is how you get back to X again this
1:32:59
question right here is to show you guys how to prove this right here right all right so to handle this well we do know see Y is equal to X which is the same as saying X over one and by looking at this right here we can come up with a right triangle so I'll will put down a right triangle right here and the reason to do so is because we can put the angle here that's why and we can label the sides and remember secant is what secant is hypotenuse over adjacent when
1:33:36
you're talking about right triangle so that means X is right here and Y is Right Here and Now for the other side that's missing well the easier way to find this out is just you open the square root and you square the hypotenuse which is X square and you minus the square of this side which is one square by looking at this triangle you can figure whatever you want of course for tangent as well here is the deal in the end you will see we have a one on the top over secant Y is
1:34:09
just X over one which is just X that's good and tangent of Y of tangent of this picture here is just the opposite which is this over over the adjacent over one doesn't matter so it's just that so we can just put down square root of x square and then minus one square which is the same as just minus one that's it that's that's it so this right here is how you prove the inverse secant has the derivative being that right so this is one of the ways to prove it I will
1:34:49
show you guys how to prove the inverse tangent derivative by using the definition once we get here all right number 23 now was question number 23 man when can I drink the ninja milk I don't know
1:35:26
that's I like to tell you guys this right here is seriously just running a mirone so this is like halfway almost halfway done so M 6 each right that was question number 23 and now moving on to question number 24 of course 24 and we have parentheses x - Y 2 and that's equal to sin x + sin y like that all right implicit differenti action in action so D DX here we go right here bring the two to the front because you have to look at the big picture so power rule
1:36:13
first side bring the two to the front and the minus one so here we have two open the parenthesis and then the inside States the same x - Y and do the CH do look at the inside and ask yourself what the derivative is derivative X is 1 minus the derivative of y is dy DX and notice that it has two terms so you have to put a parenthesis that's it for the left hand side and now for the right hand side this right here you don't need to do the product rule just
1:36:46
differentiate the first which is going to give us cosine X and different differ the second we get plus cosine y but the derivative Y is dy DX so don't forget that dydx right here just like that H well we have to multiply this out and based on how I come up with the yeah just just multiply this out anyway I think I I multiply this out as well so I will just multip I'll do this for you guys this and enough first so
1:37:23
I'll multiply this out this way for you guys when we do that I will keep the 2xy 2 * x - y right I'll take this quantity times one which is still that and then minus 2 * x - Y and we have the dydx right and that it's equal to the right hand side which the right hand side didn't do a thing so I just write it down cosine X plus cosine Y and again the Dy DX right here okay I would like to keep the dydx on the right hand side so that I can make that positive and I'll bring that to
1:38:07
the other side so perhaps I'll just write it down right here first this right here stays so it's 2 * x - y bring this to the other side which is minus cosine X and that's equal to well
1:38:22
bring this to the other side it becomes positive2 parentheses x - Y and we add this with cosine of Y and then both of them have the dydx so Factor the out put it at the end like that right and in the end of course we can just bring this to The Other Side by dividing and that's pretty much it so we'll do that finally Dy DX the answer to that is this on the top but of course now it's a time to distribute so we have 2X minus 2 Y and then minus cosine X and bring this to the
1:39:03
other side right so I also divide that this will be on the bottom 2x - 2 y plus cosine y so this right here is a plus right this right here is a plus that's it this is right here is the answer that was question number 24 and I think it's a fair time for me to change the marker this time unfortunately I cannot I cannot uh do 100 derivatives with one marker I don't have a fresh marker hopefully this one will work well ah much better okay that was question number 24
1:39:53
okay okay okay okay okay okay okay okay oh look at that look at that next one we are going to find dydx for the following x to the Y's power equals y to the X power this is actually one of the really really famous equ this is I mean the equation wise why because I have a feed on this equation how to solve for x and y that satisfy this equation and has a lot of use anyway we are going to find the derivative right here well this is the function why it's also a function so we
1:40:36
cannot just uh do anything in the moment earlier with did X to thex and then we rot the base X as e to the natural log of x that'll show you guys this way right here right we can do that we can write X as e to the natural log of x raised to the Y and we can write y as e to the natural log of Y and then rais to the X we can still do that but in this case it might be easier if we just take the
1:41:06
natural log on both sides to begin with we would like to do some algebra first and this is also a pretty good strategy whenever you have a function to a function power or maybe maybe when you have a lot of functions multiplying together while dividing you can take the natural on both sides you can use the natural properties to bring things down anyway in up talking here's the deal this is
1:41:32
y as a power we can bring the Y to the front but do not do not do not minus one because we are not doing derivative this is just the natural property bring the Y to the front we have y * natural log of x like this okay and then similarly we bring this to the front we have x * natural log of Y seriously just like that okay now we are going to differentiate so put the DDX right here it doesn't have any weird situations this is just the innocent case you can just use the product rule
1:42:09
now so here we go product rule in action we keep the first function which is y * the derivative the second derivative natural log of x is just 1 /x and then we add the second function natural log of x times the derivative of the first derivative of y is dydx here is the dydx AHA then we add what mean we have this equal to then we have this equal to here we have the product situation again keep the first function which is x times the derivative the second derivative of natural
1:42:44
log Y is 1 / y * the derivative y thanks to chend do here is the dydx again right and lastly we have to add the second function which is natural log of Y time the derivative of the first which is just a nice one right there okay so that's the idea well well right here again we see that we have an equation with fractions so we have the X and Y so perhaps let's multiply Everybody by this denominator and that denominator let's put down the X Y all the way the end okay so you guys can
1:43:22
still see this one anyway here we go just draw arrows for you guys X cancel by Y and Y becomes y s plus this times that so it's x y and then natural log of x and then Dy DX this times that the Y cancel so we have X squ so this is equal to x² and this is dy DX and this is X y oops X Y natural log of Y like that you know back here now move this to the other side move that to the other side so we have this right here I can just keep it X Y natural log of x hold on let me see
1:44:08
how I subtract yeah I I did it this way because that's why I put down for the answer choices of course if you subtract the other way that's okay too and then you just get the backwards version and then it's actually equivalent so that doesn't matter anyway I'll move this to the other side so we have Min - x s and we of course factor out the dydx put it at the end and this is equal to we have this right here first which is X Y natural log of Y and then minus y s like that
1:44:42
in the end divide this on both sides and we'll be done so finally Dy DX equals XY natural log of Y y - y^ 2 all over X Y natural log of x - x² it has a pretty nice symmetry in my opinion just like that very very nice isn't it yeah first page done look at that look at that look at that so page four actually but the first sheet done man so I'm afraid man this has a lot it's okay okay it's
1:45:22
okay it's okay it's like seriously for the first one for the first um integral of 100 integrals I didn't know if I was able to do it or not just like my first ever Marathon I was not sure if I was able to finish it or not but I did just like I did finish the 100 integrals right and after I finished that it's like man I just want to do more and more and more it's just like I've been running about I I have done Nine mirror songs in my life already it's not a lot compared to the
1:46:00
elite Runners but it's a lot for me seriously because I'm not like good athlete in the first place so I'm enjoying it a lot anyway number 26 we are going to find the derivative for iners tangent and the input is x * y X2 * Y and that's equal to x + y to third power all right so this right here it's not so bad I don't recommend you guys to take the original tangent on both sides though we'll just keep this right here because the derivative in the end will be more
1:46:43
algebraic greater than you'll see like inverse tension whatsoever I mean you know what I mean anyway go ahead and do the derivative here's the deal inverse tangent right here we get one over 1 plus parentheses you put the input right here and square that all the time okay so you have 1/ 1+ parentheses x² * Y and we have the square that this is just a general form for inverse tangent when you take the derivative of that but we have to do the CH
1:47:27
do the derivative the inside so here we go derivative of x * y so x s you keep it and then you differentiate the second so that will be Dy DX that's the derivative Y and we add the second function which is y times the derivative of the first which is just 2x just like that not so bad huh now on the right hand side that's the even better news because it's not so bad all derivative X is one and you don't do product rule here it's just a sum so you just
1:47:58
do the derivative X which is one and you add derivative of y to the 3 power is 3 Y 2 d y DX just like that okay again we have equation with fraction and I did clear the fraction before I go so uh we'll see this right here if you look at and just work this out this is 1 + x to the four power y to the second power right so I would prefer to just multiply everybody right here by that denominator which is 1 + x^ 4th power Y2 okay and again let me just show you guys we
1:48:45
multiply this right there and this right here and this right here I only draw three arrows because this quantity is being multiplied by that when we do this times that this and that cancel so we just have this or another thing is you can just think about multiply this on both sides yeah anyway so on the left hand side we get x² Dy DX and then 2xy so let's just put that down plus 2xy and that's
1:49:19
equal to I will do this times one which is just uh one plus let me just put it down like this in blue 1 * that so we can just have 1 + 4 x sorry 1 + X4 x to the 4 y s right one times that then what we do next is we do this times that but it has the Dy DX so I'm not not going to multiply the out yet I'll just keep it as plus this is 3 y Square times this thing actually that's that's that's do this 3 y^ 2are * 1 is 3 Y 2 and then 3 y s * that is + 3 and we have the X to
1:50:16
the 4 power y to the 4 power like this right I take this and multiply by that and then we have the dydx all the way at the end and notice I didn't distribute the dydx okay for a good reason you can do it but doesn't really matter let me see it's Al it's actually a good idea it's let's just do that let's just do that why not all right so multiply this and that so we have three y^ 2 dydx * y so 3 y^ 2 and perhaps I'll just put on
1:50:54
dydx right here and then this times that we have plus 3x4 y to 4 Dy DX all right so this is what we have all right and that's this is it for the for the calculus and now this are just the algebra part anyway I am going to move this this to the other side factored out the dydx so you see see here I'll open this right here is X squ it has the dydx I will Factor it out and put it at the end I
1:51:26
will move this to the other side so it becomes 3 y^2 and the dydx will be happening at the end and I'll also put this to the other side that will be minus and we will have three x to the 4 power y to the 4th power oh all in all everybody will have the dydx so we'll just put down the dydx at the end like this and then we have this this and that so we just have 1 + x^ 4 y the 2 bring this to the other side it becomes Min - 2 XY yeah that's pretty much it and then divide this to the
1:52:07
other side and we are done so finally Dy DX = 1 + x 4 power y^ 2 - 2x y all over that guy X2 - 3 Y 2 - 3x4 y to 4 like this and I would like to check my answer choice to see if I can find it ah I have the other order again so this is the travel like depends on how you want to no the way I did it is I'm move this to the other side so that's why you see the answer key is that's why you see the answer Choice has the X squ be
1:53:01
negative anyway this is the part that you just have to kind of work out the um sign on your own the way I did this is I put this to the other side and I put this I put this two on the other side so the answer key okay or you can just write this down Dy DX let's try to see if we can do this in our head um I maintain this now no oh I bring this to the other side so I have 2X y being negative
1:53:39
2x y being sorry 2x y being positive I move this and that to the other side so I have minus1 and then minus X4 y to the 2 and then divided by I kept this and that positive so we have 3x² and plus 3x4 y to the 4 and then minus I'll bring that to the other side so that's minus x to the second power so that's pretty much it it's the same as that but let just all by just multiply negative negative you get that so depends on how you uh how you move things around so I box this
1:54:14
because that's the one that I have on the answer key on the answer choices I say anyway let's see all right number 27 here we have x² / X2 - y s and that's equal to 3 Y how many of you guys ready to do the question rule on the left hand side raise your hand good nobody because because I don't like it either this is what we do multiply both sides by this denominator x² - Y 2 why this times that will be gone we will get X squ and that will be equal to this times that just multiply out
1:55:05
we don't have to do the quo rule if we don't need to so when we do that you'll see this and that together we will get three x² Y and this and that together will give us minus 3 y to the third power this equation and that equations are the same well of course you have to worry about the domain but derivative right here just focus on the D derivative anyway right here we differentiate so DDX in action on the left hand side this is everybody's favorite derivative equation it's just
1:55:37
2x on the right hand side okay here we go product rule in action first function keep it multiply by the Der to the second derivative of y is dy DX so just put that down plus the second function which is y times the derivative of the first which is 6 x like that right 6X well right here we have to bring the to the front minus one and we have Min - 9 Y 2 and remember Y is a function of X so we have
1:56:11
to multiply by Dy DX like that okay so this part I just of course keep this and that on the left on the right hand side move this to the other side so we get 2x this to the other side becomes minus 6X Y and this right here is equal to I can factor out the dydx I will write down 3x 2 - 9 y^ 2 and dydx okay all right divide this on the other side so we get dydx keep in mind this is on the top this is right here 2x - 6X y all over this right here 3x² - 9 y to the 2
1:56:59
power can we simplify uh I think so what the I not do it no we cannot simplify uh on the top we can f two but doesn't do us any good I didn't I didn't Factor Anything yeah just leave it like that because you factor it out we cannot cancel so why bother right so that's it all right now I would like to do number 28 side by side with number 27 you will see why okay here we go number 28 we have e to the natural log of x minus so e to the X over Y and that's equal to x + y
1:57:59
s all right first thing first you can take the natural L on both sides and get R of the e that's fine but doesn't really matter you can just deal with the e that's that's just as fine okay so I would just like to do the question as out is notice that we have X over y with that we have to just use the quo rule if you take the natural log on both sides you can multiply by Y and let me know how that goes you have to use the CH rule a lot so I think that's a tradeoff
1:58:32
but up to you anyway here D DX all the way across left hand side e to the XY the derivative of that we just keep it first which is e to the X over Y and then multiply by the derivative of the inside and we have to use the quo rule here we go Square the bottom function which is y square like this right so Square the bottom function and then we bring down the we write down the bottom function times the derivative of the top derivative of x is one and then we minus
1:59:13
the top function which is X and we multiply by the derivative of the bottom function der of Y is dy DX so it's like that done for the left hand side and then for the right hand side this is nice if you are willing to just keep the natural keep the E2 on the left hand side right if you don't do the natural so here derivative of x is one and derivative of y to the second power is 2 Y and the Dy DX like that
1:59:50
okay now let's do the following because again we have an equation with fractions right here so it's a better idea to just clear out the fractions right now let's multiply Everybody by the denominator which is just a y s so I will multiply this with the whole thing here and then multiply and multiply yeah just like that okay so if you look at this part this times that will cancel out
2:00:24
but we still have e to the X over y times this and that let's distribute it this into here this into here so first we will have y time e to the X over y right and then we have minus X and then let's write that down first e to the X over Y and then let's put down the dydx usually have the dydx at the end that's a good idea okay that's that's it and then on the right hand side y sare * that is y^ sare y sare * that is plus 2 Y3 power and then the Dy d x okay to make things positive
2:01:05
let's bring this to the other side let's bring this to the other side so we get y e to the X over y minus y 2 and this is equal to okay I will put this down first parentheses 2 y 3 power bring this down becomes positive right to the other side positive x e to the X over Y and then both of them have X both of them have Dy DX so I just put the dydx all the way at the end like that okay now divide this guy on both sides and will be done so we will have the Dy DX equal to here we have
2:01:51
the y e x over y power - y^ 2 over well I think I like to put a similar term in the front right so I did have this right here go first I just want to show you guys how I have the answer ke I have this first so I will just put this down x e to the X over y Plus to Y to the 3 power like that okay and of course if you switch W on the bottom nobody will get mad at you just kind of pick the
2:02:27
best answer on answer choices so that one is the implicit differentiation with quo rule this right here you should not do the equation rule okay so that was 28 and two more I can drink my P milk milk when I first saw his video video that he said he had a energy drink I thought he was joking because he always does that right but then I watched the whole thing oh my God it was legit so I wed it right away because you know I used to watch him a lot when I was probably your age
2:03:07
yeah right so subscribe to nikaha that's his Channel's name [ __ ] G um hi G I'll have the link in the description why not okay number 29 number 29 x² + y s minus one to the 3 power and this is equal to Y like this okay so let's just deal with this and the good thing is that I didn't multiply out anything for the answer key you will see what I mean by that so let's go ahead and differentiate on both sides okay
2:04:02
here we go on the left hand side look at this to the third power so bring the three to the front and then minus one so here we go we have three parentheses the inside stat the same X2 + y^ 2 - 1 to the 2 power and then we mulp mly by the derivative the inside open the parenthesis because we have more terms coming okay derivative x s is 2x plus the derivative of y^ s is 2 y Dy DX and then the derivative of minus one is just zero so we are done yeah
2:04:38
that's it on the right hand side we have this the derivative of y with respect to X is just dydx and you see hopefully the notation makes sense DDX of Y is just dydx okay now simplify this guy this is what I mean I didn't multiply this out we just keep this okay so to cling things up we will take this as the whole thing multiply with this multiply with that so let's distribute it first so 2x * 3 is 6X and then we have all that which is X 2 + y^ 2 - 1 2 yeah and
2:05:24
then 3 * 2 Y is + 6 Y and then of course we write that down first before the dydx so let's put that down right here and dydx and right hand side we still have the dydx as well okay okay you know the deal bring this to the other side so you have one and all that so perhaps I'll show you guys more work 6X x² I know some of you guys can do this in your head but I to show you guys all the work here right hand side this is like saying one dydx so we have a one and then this on the
2:06:04
other side becomes minus 6 y parentheses X2 + y^ 2 - 1 square and then everybody had a dydx right here so I the out now bring this to The Other Side by dividing and that's it so Dy DX equals that's on the top so just do that 6 x X2 + y^ 2 - 1 2 or over 1 - 6 y parentheses X2 + y^ 2 - 1 2 like this done okay good number 30 my my God finally but it's not any easier why because we are doing second derivative from number 30 to number 39 why anyway this right here I will
2:07:04
have to write down the question for you guys we are going to find d2y dx2 and if you look at this right here this is the second derivative notation the idea is because well to get the second derivative you just have to differentiate the first derivative right well as I said we have to differentiate this is the notation for differentiating we have to differentiate the first derivative oops Dy DX we have to differentiate the first derivative
2:07:40
and now you see we have D and D so that's D squ that's just a notation it's not really d square but we put on D2 y That's how you say it D2 Y and on the bottom you see you have DX and DX happening twice so we have dx2 that's the notation for derivative for the second derivative okay all right what's the equation that's the main thing huh we have 9x2 + y^ 2 equals 9 okay here we go we have to differentiate this twice why it's not isolated
2:08:15
it's okay just do implicit differentiation go ahead DDX do now on the left hand side this is 18 right 18 x to the first Power Dera this and the derivative that is plus 2 y y is a function of X so we have the dydx like that well the derivative of 9 with respect to X is zero so this is equal to zero good now check this out I will have to move this to the other side so we have 2 y Dy DX and
2:08:50
that's equal to this to the other side becomes 8X and we will just divide the 2 y on both sides so they cancel and you just get Dy DX equals reduce that you get Negative X on the top 9x on the top right 9x on the top over y Sox over Y and you see this right here is the first derivative we just have to look at this expression and differentiate that again so we can get a D2 y dx2 so let's do
2:09:26
that right here let just emphasize that right d DX this right here right d DX this right here so we see right here we get d2y dx2 on the right hand side we have to do the equation rule again just like the example that we did number 28 believe Square the denominator so parenthesis y square that okay the bottom function is y times the derivative of the top the Der 9x is just 9 so this is the the on the top minus the top function which is -9x top function just write it down times
2:10:13
the derivative of the bottom the derivative of y is what Dy DX isn't it I'll put this down der Y is dy DX okay so you see that the second derivative it's based on the first derivative and other stuff right but the truth is if you look back Dy DX look back right here this right here is just -9x over y so we can replace this right here real quick so we see for the second derivative we will end up let's see we end up 9 y negative negative without any further
2:10:55
explanation just positive 9x right here and we can replace that we can replace the dydx with this so I will do that real quick wait I thought I turn this thing off already it's the Wi-Fi that want me to connect no I don't want that see so y5 telling me to update okay anyway this right here just put it down right here which is --9x over y okay and we divide this by y square like this so far so
2:11:36
good right hopefully that's the case and of course again this right here we had a complex fraction so perhaps we do the following so the rest is just some kind of a simp simplification okay here we go I multiply the time and bound by y because so that I can get R of the complex fraction situation and uh let's see I should be able to fit in here let's see this is going to give us negative 9 y s
2:12:09
and this is going to give us Y and Y cancel out but 9 * 9 is 81 and we have the x s yeah over this right here is y to the 3 power like this okay and the moment that you think hey we are done but we are not because in fact we can Factor something out on the top and things will work out pretty nicely on the top we can factor out -9 and you see if we do that here we get y square and we add and here we get 9 x² like this we fact out and does this look forar 9 x² + y s check this
2:12:55
out 9 X2 + Y 2 this right here is just equal to 9 so all this right here is just equal to nine this right here is equal to nine of course the water doesn't matter of the adding so this is just equal to nine therefore finally you can simplify the expression to
2:13:19
81 over y to the 3 power and this right here is it okay this right here is it oh man this is number 30 so that means I have three more to go this is going to take more than 6 hours for sure why do I do do this to myself this is going to be more than six hours for sure anyway let me haveh the Ninja milk ninja milk so this is 2 hour and 14 minutes uh give this a time stamp right here man oh so this is how it looks like so it's like cical like the the Japanese
2:14:21
drink and it's sparkling this is a sparkling can you guys see hey it's pretty good actually it tastes like Cal Co I think that's how you say it um the Japanese drink and this is sparkling and this is the organic milk with flavor so take a look right here Ryan Higa n Nika's ninja milk energy drink I don't know if this will really give me the energy to last for another I don't know how many hours really good
2:15:20
wow tastes really good actually 112 calories per can right so for the people if you care about nutrition here this is the nutrition label I'll drink more how many caffeine do I have I don't have any caffeine here oh caffeine's 80 millgram you can see right here caffine 80 8 migr right here and I give one gram of protein look at that oh man I I feel good tastes really good W okay no time to rest though that was just uh as I
2:16:11
said question number 30 and I will drink the ninja milk tastes good good tastes good okay question number 31 well we just have to do second derivatives I serious the D is my clock okay man if I have to do four more hours I'll be going home around 10 let me see if I can go home by 10:00 I don't know okay if I shut up and do it I will okay number 31 right here all right differentiating actually no D2 dx2 meaning differentiate this twice okay D2 dx2 D2 dx2
2:17:13
differentiating whatever is inside twice here we have 1/ 9 secant to the secant of 3x just like that okay so here is the D D2 dx2 means DX sorry DDX DDX okay so I'll do this one right here just to show you guys what we have to do we have to differentiate of differentiating like this and then 1 over 9 secant of 3x so we look at this do it inside out so here we go and perhaps for the inside we just do the end do the part in black I mean in blue and then the final answer I'll
2:18:01
end it with black okay so differentiate this part okay this is not so bad we have one over n it's just a constant multiple write it down write it down write it down now differentiating secant of whatever you put down secant of that whatever time t of the same but we have to multiply by the derivative of that not me multiply by three so that's pretty much the idea well we can reduce this that's just one over three so we will this is going to be differentiating 1 over three and
2:18:39
here we have secant of 3x * tangent of 3x now differentiating this product R in action this will be my first function I'll keep it which is 1/ 3 secant of 3x times the derivative the second derivative tangent is secant thir secant to the second power and the input is 3x times derivative of three the Der 3x which is three because of the chandus as so okay so this is the derivative of tangent of 3x now we have to add the second function which is tangent of
2:19:23
3x times the derivative of the first derivative of 13 of secant to secant 3x you write down 1/3 secant actually oh I messed up with the color sorry let me do it again I messed up with the color Okay first function time C der second Dera der tangent is secant second power of 3x but we multiply by the derivative of the inside which is the three right here now we add the second function which is tangent of 3x then we multiply by the derivative of the first
2:20:03
which is the 1/3 and the derivative secant is secant of the input time tangent of the input and the derivative of that input so you multiply by three right here so that's pretty much it and now you see that 1 thir and three cancel for the first part and this one and this three cancel for the second part and we just put things together finally the answer is secant to the third power right this times that first Power second power third power and here we have the 3x like this
2:20:42
and we add this is secant let's just write down the secant first secant 3x and tangent tangent so that's tangent second power right second power 3x and again you can perhaps fact out the secant of 3x but secant square plus tangent Square not so much you can do with it so just leave it like that in fact that's what I did as well that's the answer okay this right here is the answer for it all right uh what's F to 31 of course 32 but it's just another derivative
2:21:26
second derivative number 32 okay D2 dx2 of we have x + 1 over s otk of X okay do we need to use question rule no we don't because thankfully the bottom is just have one term so that means we can split the fraction nicely so we will just have to differentiate right the second derivative okay X over square root of x and then plus 1/ sare root of x remember remember remember the square root is like the same as what one 12 power so perhaps I should have write this
2:22:11
as the I should have written this down as the 1 half power so right here let's put on 1 half put on2 if you would like well well that means what differentiating this twice this is the saying x to the first power over x to the second power so subtract the powers we get just 1/2 so we have x to the positive 1/2 power for the first one for the second one we have x to the negative2 okay and now here we go notice I just put on D DX because I'm going to do the first derivative right
2:22:49
here already let's bring the power to the front minus one bring the power to the front and minus one so for the inside here you'll see the first answer here is2 and you have X to thetive one2 and let's just write it down as a power because we have to do the derivative again similarly for the second one we have negative though negative one2 and then x to the -2 minus 1 is 32 oo I don't know why I did that but whatever and then finally you bring this to the front
2:23:33
and do the derivative again right I don't know why I thought it was a good idea to just show you guys a lot of derivatives like even the second derivative so I'm actually doing like like I don't know 20 derivative right here I don't know anyway this times that no more DDX however just the answer this times that is -1 over4 and this times that don't forget minus one don't forget minus one so we get x to the 32 negative negative becomes plus
2:24:05
three over 2 * 2 on the bottom so we have over four and then this is X -32 minus 1 is 5 over 2 like that okay you think you are done but no not yet bring the X down to the denominator and then work out the simplification so you get -1 over 4 and this is X to the 3 over 2 power and plus 3 over 4 and this is X to the 5 over 2 power can we get some common denominator and clean things up yeah well this is three2 if I can get five2 this is the same as say x to2 over2 which
2:24:55
is just x to the first Power I'll put on as X to 2 over two that's all we need right well all in all you get four x to the 52 and I didn't write it as a radical for the answer choices we just keep the answer like this and this right here is negative X plus three and I believe that it's it and I even have the well I didn't have the right order but I put on 3 minus X which is the same thing okay so of course perhaps I just right this 3 - x over 4X 5/ 2 power like this good
2:25:48
man this is so computational like for the derivatives they are just so computational you have to do a lot a lot of algebra and I don't know why like I I was trying to come with the questions right huh maybe the question is too easy so I make it harder so I was just having like 100 first derivative so let me make it harder let me put in some implicit differentiation I did let me put in some second derivative I did let me put in some definition of derivative I did
2:26:24
man okay number 33 here we go D DX and we are going to have well sorry I my dream D2 over dx2 right here we have inverse s of X squ oh my goodness inverse s of x s of course here is everybody's note right everybody's favorite note okay to differentiate inverse sin x this right here is 1 / < TK 1 - X2 and maybe you guys can use the chance to kind of Google it to see if I did it right or not but just help me out if you can tell
2:27:18
me if it's WR or not anyway it's not funny I know okay here we go uh here is the input but you see I will actually just write down D DX okay because I'm doing one theity already in first sign I get one over open the square root right open the square root and then one minus the input and square the input is x square but you have to square this again like that are we done no we still have to use the chandle multiply by the derivative
2:27:56
inside derivative of x s is 2x so that's what we have okay man we have to differentiate what now differentiating 2x over sare root of 1 - x to the 4th power oh man quotient rule
2:28:19
put down the denominator square that so we have square root of 1 - x to the 4th power and we square that bottom function put it on the top which is square otk of 1 - x 4 power time the derivative of the top derivative of 2x is just a two and then we minus the top function which is 2X and we multiply by the derivative of the bottom it's a square root so we first have 1 over two square root I have one keep that in mind one over uh two
2:28:53
square root of the inside which is 1 - x^ the 4th power but don't forget the Chanel says we have to multiply by the derivative of the top of the inside derivative of negative 1us x 4 power we get -4x to the 3 power like this okay just like that okay okay so cancel yes but complex fraction we have this guy in the denominator on the top no good so let's multiply the top and bottom by the square root of 1 - x to four power okay yeah all right so multiply multiply and this
2:29:44
are gone already the square and the square root so they are gone already um so let's see what we have okay square root square root they cancel when they multiply so you have 1 minus x four power but you have to multiply by two so you actually get two * theide which is two 2 * that's minus 2x to four power okay and then let's see this and that cancel out as well okay two cancel out it's this two yeah and um negative * negative is positive four and
2:30:26
x to the fourth power yes over how do I write the answer key okay I have the third okay okay okay so this right here I put it as the first power so I write it one - x to fourth power to the first Power this right here R is 1 - x to the sorry this should have been a four this and that should be the same so 1 - x to the four power in the square root so this right here is 1 - x to four power to the 1/2 power okay on the top we can reduce that 2 minus not plus this and
2:31:05
that together we can get two x to the 4th power over now for the bottom they have the same base just add the exponent so we get 1 minus x to the fourth power to the three half power but I change that to a square root form so I just give you guys a square root answer which is right here you have the over two that becomes the radical the square root so in the end in the end in the end two x to the 4th power over this right here will give us a
2:31:44
square root and we have 1 - x to the 4 power parentheses that and raise that to the third power that's the answer this is the answer that's it number 34 D2 dx2 of 1 over 1 + cosine X like this quotient rule or not no quo rule good okay
2:32:17
let's rewrite this as a negative power first let's put it here D2 dx2 We can write it as 1 1 + cosine X to thetive 1 power like this so let's do the first derivative oh man I forgot to do the first derivative in blue over there I'm sorry but know it's okay I'll do the first derivative in blue right here the inside here will be the first derivative right here I will first bring the negative 1 to the front FR and then minus one so we get Negative parenthesis the inside
2:32:52
States the same which is one plus cosine X to the -2 power and then we multiply by the Der inside which is sinx that's the derivative of the inside like that right can we clan things up yes we have DDX negative negative becomes positive positive sin x over 1 + cosine X raised to the 2 power like this now we really have to do the question rule you might want to do the product rule right here but um depends depends so here we go I I will just give myself more
2:33:39
space Square the denominator so let me put this down one + cosine x² but we will have to raise that to the second power put down the denominator right here which is 1 + cosine x square don't forget a square times the derivative of the top derivative of s is cosine and then minus the top function which is sin x times the derivative of the bottom this right here bring the power to the front so we multiply by two parenthesis 1 + cosine X to the first Power and then multiply by
2:34:21
the derivative inside which will give us sin x like that oh man let's see let's see how do you simplify that I did simplify the top so we have to simplify the top oh actually I see how to simplify the top here you don't have to really multiply these things out because we see this and that can be F out so we'll do that first can you see the Factor tonight can you see the Factor tonight okay
2:35:00
right here we have the first Power right so we have 1 + cosine X but don't forget we still have this cosine X good this is out negative * negative is positive so so plus just emphasize plus you have the two and then you have the S Square X like that yes good over this guy is 1 + cosine X to the four power on all now this and that can be cancel we have three more left and uh that's pretty much it on the top we just distribute this and that's cosine square and this is two
2:35:44
sin Square so we can combine anything so finally cosine X plus cosine squ X plus 2 sin s x all over this guy which is 1+ cosine X to the 3 power and then we are done this right here is the answer answer why do I make the question so hard all right that's 34 right that's 34
2:36:49
okay 35 question number 35 second derivative x times the inverse tangent of X go ahead and do it D DX put this down in blue though first function which is x times the Dera second that give us 1/ 1 + x² and we add the second function which is inverse tangent times the derivative of the first which is just a one so that's what we have nothing too special all right now we have to differentiate
2:37:33
this right here well we can just put the X on the top right we haven't took we haven't taken the derivative for the for another time so we just put the X on the top and then this is just the inverse tangent of x well just clean up so this is the first derivative now just do it for the first one let's use the equation rule so Square the bottom so we will have 1 + x² and we square that then the bottom function goes to the top 1 + x² time the derivative of the
2:38:09
top and then minus the top function which is x times the derivative of the bottom which is going to give us 2X and then we add the derivative of infers tangent which is 1/ 1 + x² like this now we are going to do some like little engineering that kind of thing not so much though it's just like look at this part look at that part com by like turns okay I see that this right here it's just 1 + x² * 1 just that minus this time that is just 2 2 x² that's good but perhaps let's
2:38:49
get some common denominator so I will multiply this by 1 + x² and also 1 + x² well thankfully this is just plus so we can just add everybody on the top so this is just + 1 * this is just one just all that so that's pretty nice let's see let's see let's see what we get at the end parentheses 1 + x squ and then square that for the bottom hm check this out x square x square 2x s and then minus 2x s oh my God cancel this out and on the top we just have to do 1 +
2:39:31
1 I think it's two yeah so that's the answer I believe let me know if I get this right or not so oh my God that simplifies is pretty nicely number 36 number 36 we have second derivative of x to the 4 power natural log of x just like this here we go okay for the first one we have to use the product rule right so perhaps here D DX parentheses product Ru action keep the first function let see what's the
2:40:14
next one uh yeah let's just first function X4 * the derivative the second which is 1 /x and um this second I'm not editing this video Steve okay plus um natural log of x times the derivative of the first which is 4 x to the 3 power like this okay good now of course this right here is just 3x X to power right so on all we are looking at to differentiate this guy together will be simplify to x to the 3 power and this guy oh my God plus 4 x Cub * natural log
2:41:08
of x derivative again why not derivative this guy is 3x² and we add the Der of this guy this is the first function 4x^ 3 power time the Der this right 1 /x plus the second function which is natural log of x times the derivative of this guy which is 12 x² like this okay of course this part again we can nicely simplify it so it's just two 3 + 4 is 7 and x² right so this is 7 x² put the 12 x²
2:41:51
first so 12 x² and then we have the natural L of X and that's it nothing else nothing else nothing else yeah be really careful though if you look at my equation number 36 there's no typo but uh I just make a small change for the answer key so aha if you are taking AP test let me know you know how the AP test the multiple choice questions are they are really tricky because this because the answer choices are very very similar okay 37 D2 dx2 e raised to the ne x²
2:42:39
power one of my favorite function here we go we have to differentiate the first derivative and for the first derivative we have this right here stays which is just e to thex S but Chu says we have to multiply by the Der inside derivative of that is negative 2x like this right so on all we just have to differentiate let me just write it down again just have to differentiate -2X * e to thex 2 so let's do it first function you keep it times the derivative the second derivative this
2:43:24
is e to thex S and we multiply by the D of that because of the chandu so derivative to that is -2X I don't even know if I'm saying derivative correctly or not but what anyway we add the second function which is e to thex stimes the the derivative of the first which is -2 all right finally we clean things up this and that give us 4 x² and uh e to the x² and this is min - 2 e to thex S and some people may want me to factor oh actually I did factor that myself too so I will
2:44:10
factor out because this right here can be Factor it so we will have parenthesis 4x2 - 2 and uh e to the x² like this this is preferred it if you need to set this equal to zero and find thetical numbers so oh man I cannot say this is it but this is it for number 37 man this is
2:44:50
okay number 38 and 39 I will do them side by side so here is number [Music] 38 so D2 dx2 for the first one we have cosine of natural log of x and in fact you can do the integral of this you can you can also do the inte of that not just the derivative and then I will do the 39 on the side this is D2 over dx2 natural log of cosine of x like this okay we have two derivative equations on the spot they look similar but of course they're also very different let's look at
2:45:34
this one first okay differentiating well we have to get the first derivative right here derivative of cosine of something first we get negative sign and the input stats the same and then what we have to do the chle the derivative of the inside which is 1 /x so I'll put on overx like this yeah overx now differentiate that guy question rule okay here we go Square the bottom which is X and we square that X and then we square that and I will put down the button function right here
2:46:22
times the derivative of the top this is negative so I will just keep that negative and we have the derivative of sign do that you get cosine and the input States the same again but we have to multiply by the derivative of the inside right here which is going to give us one /x so that's it for the first part then we subtract the top function and the top function has negative so you write down negative s of natural log of x okay this is top function and then we multiply by the
2:47:00
derivative of the bottom and the derivative of x is just one so that's what we have and check this out X and then this is overx so this and that cancel out very nicely and you will see that this is negative cosine of natural log of x and negative * negative is positive s of natural log of x all over x² all over x² and on the answer key I put down the sign first because I put on sign over that but I'll leave that to you guys okay so just do that now for this one here we go let's
2:47:45
get the first derivative right here inside I'll do that in blue this time though you have to remember when you differentiate natural log you first do one over whatever this is which is cosine X and then the CH says multiply by the derivative of the inside derivative of cosine X is what yes negative sin x like this good so check this out this is going to be just to take the Der ative of negative
2:48:17
sin x over cosine X this right here is nicely equal to tangent X well the derivative of negative tangent X is negative you have secant Square X that's it which one did you guys like more see you just switch the water of the input and the uh input and for the in function alter function say oh my God that was number 39 I don't have to do like two derivative for one question again man good number 40 here we go ah this is a very good one D DX uh I need more
2:49:02
space negative SEC squ X number 40 d DX DDX it's just one derivative okay Square Ro TK of 1 - x s and we add x * INF sign of X here we go just one derivative of the square root we get one over two square root of the bottom of the inside like this on the bottom like that right and we multiply by the derivative of the inside so we multiply by -2X that's it for the first derivative for this part and then we add this part requires the product rule that's why
2:49:54
I have the dot keep the first function times the derivative the second derivative inverse sign is 1 over Square < TK of 1 - x 2 H similar and then we add the second function which is inverse s of x times the derivative of x the the first function now check this out cancel cancel so if you write this down again you'll see this is negative X over this is sare < TK of 1 - x s but here is plus X over sare otk of 1 - x s and of course don't forget lastly
2:50:34
you have that inverse friend right inverse sign of X like that of course this and that can be nicely cancelled out finally you just have your inverse friend inverse s of X like this plus C no you don't have to this is it in C two you guys will know have to know how to go do this backwards the r the will give you iners sign I'm not asking you what's the derivative iners sign I'm asking you derivative what which is this will give you iners sign integration by
2:51:12
parts I don't know who are watching the videos uh my intention is of course for the students who are in C one and C 2 especially for the very last few questions or maybe you haven't done calculates for a while and you are going to do like um Cal three maybe um especially in the summer if you just want to practice you should do this maybe you are AP C students just do it well maybe you're just like me who just like derivative calculus and all that yeah number
2:51:45
41 okay just one derivative we're going to differentiate x s < TK of 4 - x² [Music] okay no just kidding I'm not liking DC all right product rule in action this is x times so here we go x times square root function so we get one over just do the Der this right here putting red 1 over 1 over 2 < TK 4 - x 2 however use the CH R multiply by the derivative of the inside and that will give us -2X that's the first part then we add the second
2:52:42
function which is square root of 4 - x stimes the derivative of the first which is just a nice one well well this two and that two can be canceled that's good and we see that x * X that's x s over this is square otk of 4 - x 2 and we have this part which is plus 4- four minus uh let me do it down this 4 - x squ in the square root right okay imagine that you had to find the derivative and you have to set the derivative equal to zero so you can find the critical numbers well this
2:53:24
is not what you preferred if you need to find if the function is increasing or decreasing or not this is not the form you prefer you want to add them together with that said we have to get the common denominator so let me just multiply the second one by that which is square root of 4 - x squ on the bottom and also on the top like this and when we do that you will see on the bottom is just square root of 4 - x 2 on the top here is x s and this times that is just plus and the
2:54:00
square root just gone we have 4 - x s like this so all in all you get four right here - x 2 - x s so that's - 2x 2 over Square < TK of 4 4 - x² like that good nice and cute huh another done 42 okay another square root differentiating square root of x squ min-1 and uh overx here we'll be using the quo rule so it's kind of like the other opposite anyway go ahead and square the denominator and put down the bottom function right here multiply
2:54:57
by the derivative of the top derivative the top is going to be 1/ 2 < TK of x² - 1 x squ minus one and then don't forget the chendu use the chendu multiply by the derivative inside which is 2X and then minus the top function which is square root of X2 - 1 times the derivative of the bottom so that's what we have on the top though again two and two cancel out nic0 and we have x s but we have a complex fraction in this situation so to handle that let's multiply
2:55:40
the top and bottom by the common denominator which is just this so multiply by square root of X2 - 1 < TK of x s - 1 and when we do that on the top this and that will cancel we have x * X which is x s and then this is going to be minus this and that the square root will be out so you just have a parenthesis and you will have x² minus one like that I haven't distributed so yeah and over here we have X squ and you have the square root of x s - 1
2:56:17
now x square - x square is just bye-bye negative negative is just positive 1 so we have one over this guy x² square otk of x s minus 1 like this okay I'm going to fit number 43 right here I I switch [Music] the order why do I do that that's interesting I just switched the W of the X over there so okay okay okay D DX so this one I have X on the top instead over square otk of x² minus 1 let's see how the answer will ter out to be okay okay question R in action
2:57:03
so here we will have to square the denominator which is this one now so put that denominator on the top right here first multiply by the derivative the top which is one minus the top function times the derivative of the bottom and the dtive butt is 1/ 2 sare < TK of x² - 1 and Chu multiply by 2x cancel cancel right that's cancel that's that's out already good however this time though again get a common denominator get get a uh um multiply this guy get rid of the
2:57:49
complex fraction so multiply the top and bottom by square root of x² minus one let's see what this will give us this times that we just have that 9 x^2 - 1 this times this part this and that cancel so we have minus x² over here this and that cancel which is that saying X2 - 1 to the first Power this right here is I saying X2 - 1 to the2 power this and that cancel on the top we have1 right over uh this right here I believe I put in this radical form for the answer and uh
2:58:35
I did so let's do this in our head 1 + 1 12 is 32 so the half is the square root and inside x² - 1 to the thir power like this that's it number 44 I'm going to f number 44 right here as well let see we are going to be differentiating this one is fun one we have cosine of inverse sinx like that okay here we go what's the derivative of cosine well it's just negative sign isn't it and the
2:59:16
input States the same so you just write it iners sign of X like this however chandu right multiply by the Der inside which is 1 / < TK 1 - x 2 just like that well well s of invers sign they pretty much cancel you can just put it down like this so you have negative x negative X over this for the denominator wow all the answer look kind of similar to each other all the square root and all that stuff very nice very nice very nice okay and uh I'm also going to fit in 45
2:59:56
here because this one is not bad at all DDX natural log of x² + 3x + 5 so this is the input and we have natural log right on the outside so remember it's one oh over the input which is x² + 3x + 5 and the Chanel says what multiply by the derivative of the inside keep in mind one over that times the derivative inside we can just put the derivative on the top so here we get 2x plus three done done man I can do the natural derivative like this all
3:00:37
day long give me 1,000 of them I will do it for you in 5 hours no don't do it don't give me 5 1,000 no no that don't take it too seriously that was me being saying whatever I don't know okay what was that aha number 46 now let do look like a six D DX this one is a tricky one inverse tangent of 4X but we raise that to the second power like this okay here is the do do not multiply this and say that's tang2 okay uh I would just say
3:01:53
don't ever write seriously don't ever write things like tangent -2 right here 4X like this don't if you want to say tangent the regular one tangent and then raised to thetive -2 power put that to the out of the parentheses okay this means you have the inverse tangent of 4X and then you square that okay don't ever write this seriously just don't I have a video on that you guys can go check that out that's like the the worst notation anyway here we go what we do is do
3:02:29
the outside which is the power situation here we have two parentheses the inside stays to the first Power and then multiply by the derivative of the inside derivative of first tangent is 1/ 1 + the input square right here right so we have to have the fourx here but don't forget we square that are we done no CH do multiply by the derivative this which is multiply by four right there now put
3:03:03
the answers together all and all 2 * 1 * 4 which is 8 and then this is also on the top so we have the inverse tangent do not bring that down to the denominator this is the arc tangent the inverse tangent so you just write it down like this over this right here is 1 plus 4 square is 16 and then you have the X squar this is like a really cool expression huh okay all right number 47 we are going to be differentiating oops here we have a cube root function by cube root of x to the
3:03:40
2 power like this do we do any CH lo you don't have to here is how you do it you just have to differentiate remember whenever you have the index here for the radical you can write that as a power so you actually get X this two is right here and then over three for the for that seriously then we can use the power rule what the power rule power L I don't know anyway here we go we get 23 x 2/3
3:04:22
minus 1 is - 1/3 like this and uh I rewrite it I brought this down to the denominator so we get 2 over 3 x to the 1/3 which is the same as cube root of x right X to 13 just like that good all right number 48 another interesting One D DX here we have S of square root of x time natural log of x wooo my question is can we find the derivative of some function that will give us this for the
3:05:02
answer namely find the integral for that I don't know anyway here we go sign is also function both of them are inside it's not sin x * Square of x times natural log of x no s is outside function times this this two inside so what we do is differentiating co differentiate s first which will get cosine and the input States the same square root of x * natural log of x like this and then multiply by the derivative of the inside okay we need a product rule in action so
3:05:39
we will do that derivative of the inside we keep the first function which is square root of x times the derivative of the second which is 1 / X plus the second function which is natural log of x times the derivative of the first which is 1 / 2 otk x like that and now we just have to simplify this expression I almost want you to say shlif but like that's does makes sense anyway hm let's see well square root of x over X we can simplify that to be just s Ro of X in the
3:06:25
denominator so I think I will just write this down again cosine of s root X and natur log X again this is just one / square root X right yeah you can do however you want to simplify that uh this is X1 12 power over x to the first Power 12 - 1 is 1/2 so you have this okay plus natural log of x over that 2 s root X ah so this is hope it's clear now we just need a two right here multiply multiply so finally the answer is do I put that down the top as
3:07:03
well I certainly did and I will do that as well I put on this first though the denominator is this which is 2 square root of x and for some reason I put on this first I think because I just want to tell you guys this is the only thing inside of the cosine so let me emphasize that by putting 2 plus natural log of x first times cosine of square root of x natural log of x like that so this is the inut for cosine in fact in differential equation you will see this kind of function
3:07:42
coine sign of square root of x * natural log of x all that stuff okay so that's good number 49 an easy one D DX of CSC not really but like you just have to do it care for the OB it's actually pretty easy actually CSC cosecant of x² so here we go differentiating cosecant we get Negative cosecant cotangent so CS C but don't forget input has to stay the same cotangent and then do what well CH do derivative of the inside 2x in the end put the 2x all the way in the front
3:08:29
-2X times all that CSC x² Co T x² done whoa look at that how easy that was right number 50 cannot fit in number 50 I will fit in number 50 right here as I said I like to fit in a lot of things onto one Bard one scratch paper that feeling is really good okay that feeling is just really really good it makes me really satisfied I know some of you guys are like me too I think so anyway here we go question rule so Square the bottom function natural log of x square
3:09:12
that guy bottom function is natural log X times the derivative of the top the derivative of the top is 2X and then minus the top function which is x² minus 1 times the derivative of the bottom which is 1 over X just like that well you can distribute it but you are going to get complex fractions so let's just multiply the top and bottom by X so we can do that right we can simplify that a little bit in the end x * this x * this guy of course we will get 2 x² and then
3:09:54
we have the natural log of x like that and then this and that cancel so you can just distribute the negative 0 so we have Min - x² and + 1 + one all over x parentheses natural log of x² done
3:10:15
okay oh by the way for the people who don't know I'm war from alpha when they say okay on war fun Alpha when they use log okay they mean natural log okay so if you are doing the your your Calculus problems on W Alpha get used to this notation anyway that was number 50 oh my God halfway done so it's going to take me about 6 hours something and I'm afraid of the last 10 questions because to be honest they were they are
3:10:50
the the they are the definition questions so I don't know how long they will take God man this is taking longer than I expect I don't know hopefully they have some easy questions coming up yes I am having some nice questions coming up number fav T1 differentiating 10 to the X now you have two ways to do it one way is just know how to differentiate any number any number okay for the base raised to the X power in that case if you remember that it's just 10 x 10 to the
3:11:32
X power 10 to the X Power Times natural log of 10 it's just always the number to the X Power Times natural log of of the base and of course you can check the CH do the derivative of x is just one so multiply by one it's the same thing this is the idea and the reason why this is true is because we can look at the 10 as the follow remember ideally speaking we like to work with space e so we can look at a 10 as e to the base I mean e to the Ln 10 this is 10 okay
3:12:10
and of course this right here multiply well rais to the X power and you can multiply the powers and you will get D DX this is e to the x * natural log of 10 like this and it's very similar to this question with did this right here you just get e to the X Ln 10 times now we have to do the Chand do the derivative of x l and 10 L and 10 is a number time x so you just SK MP by L and 10 like this and this right here is just 10 to the X power that's how you get the L and 10 okay like
3:12:55
that okay number 52 number 52 number 52 your order is ready please come to the window and pick up your what I don't know why I'm saying things like that but whatever I just did so of this yeah okay so this right here it's a cube root maybe we are not familiar with a cube root but okay we can write that as the 1/3 power so we will do that therefore we are just going to differentiate actually I will uh yeah I'll just do it here we'll differentiate inside is X Plus
3:13:48
natural log of x to the second power and then we raise this guy to the 1/3 power well power rule in action put this all the way in the front and then minus one so we see this is going to give us 13 parentheses inside States the same x + natural log of x squ what's 1 minus 1 -23 so you just have -23 here and you multiply by the derivative the inside okay yes two things so be sure you open the parentheses derivative X is 1 plus the derivative of this well we need
3:14:29
to use the power again bring the power to the front which is two inside stays the same 2 - 1 is one but CH says we have to multiply by the derivative this guy just to this though because this guy has the derivative right here so do not close the parentheses do not multiply one /x with the whole thing the one of X which is the derivative of natural log of x it's only for this part so you have to close a parenthesis like this right and how do you write the answer this is
3:15:03
not oh okay hm we have to bring this down to the denominator and just do the usual clean up because we'll end up with a complex fraction and uh I will I should have more space here okay here we go this right here is on the top so I will put it on the top 1 + 2 natural log of x * 1 /x right just bear with me right here for now over three in the bottom on the B bottom and uh this guy is
3:15:44
on the bottom negative right the three is the cube root so we have the cube root open that and this right here X Plus natural log of x this right here has a square but don't forget we still have this Square here so perhaps I will make this two red and I will Square this guy red like this well well as I said we have the 1 /x here let me multiply by the X in the front this time because if I put
3:16:18
it at the end looks like itself square root I mean the cube root but it's not anyway multiply in we get X that can also we just get plus 2 natural log of x right and uh 3x see you guys are okay huh 3x and then the cube root of x plus natural log of x this guy square and then this guy the inside guy Square as well so this right here is the answer for that okay pretty good pretty good pretty good uh as long as not like second derivative man I don't know why I did that okay number
3:17:05
53 ooh very nice just some little power fun okay Power functions so we have DDX x to the 3 4 power minus 2 x to the 1/ 4 power this right here is actually pretty common if you have to talk about how to sketch a curve because you actually get some sharp turn all that stuff so uh yeah just you know how to differentiate this and how to find a critical number for this anyway go ahead bring this to the front bring this to the front minus one minus one okay we have done this
3:17:45
question very Sim we have done the question very similar to this one so anyway we get 3 over 4 x to the 3 over 4 - 1 is -1 over4 okay -2 * 1 over 4 we get - one2 and then x2d subtract you get -3 over 4 like this okay brings in down to the denominator so here we have 3 over 4 and this is X to the POS 1 over 4 power minus 1 over 2 and this is X to the 3 over 4 power how can we get the common denominator well here I need to multiply by two so I can get a four but here I
3:18:35
need to multiply by x to the 2 over 4 so I can get x to the 3 four I know that's the same as X to2 so you know don't worry I will fix that I think it's better this way though anyway when you do that the bottom has the denominator being Just 4 x to the 3 over 4 and then for the top this is three this guy it's x to the 1/2 namely square root of x okay and then minus two like that and um of course I put down the button as a radical as well so I'll just write it
3:19:18
3 square root of why am I keep using my finger to inrease okay 3 squ Ro TK of xus 2 over 4 x to the 4 over 3 power which is the fourth root of x to the 3 power like that derivative what this this part o log base two for the next one log base two log base 2 log base two okay shouldn't be so bad because we know our log properties anyway here we are going to differentiate here is the log base 2 and we have x
3:20:04
times square root of 1+ x square like that huh but I don't know what's how to take the log base 2 derivative it's okay it's similar to this we can use log properties to take up that in the log world we like to work with natural log named dln so here is how we can do it we have DDX okay for log base 2 what we can do is just go ahead and do log we just just change just change this to Ln just change this to Ln and then here we have the inside States like this x square root
3:20:49
of and this is a multiplication 1 + x s like that but you see this is the two that means you just have to divide it by natural log of two like this okay so again natural log of whatever divide by natural log of this number is log of base of this number of the input like that okay so log properties let me just put it down under the clock this is log base B of X okay this is equal to log of x well any log works but we're talking about natural log of x over natural log of B and
3:21:30
it's just the input is right here it's slly more complicated but get used to it okay now let's do this down let's put this down right here we have natural log of two it's just a number and don't let this bother us too much so let me bring that all the way in the front that's one over natural log of two like this okay and we are going to differentiate why I open the parentheses we're going to differentiate well here if you look at the part in Black we have natural log of x times
3:22:06
natural log of square root right product in of natural log you can break them apart it becomes the sum of natural log so we can write this as natural log of x stop and close that and then plus the other natural log and this part square root of 1 + x² like this okay and of course the reason to do so is because this way you don't have to use the product rule and that's how you can actually prove the product rule as well thanks to logarithm even better though when you
3:22:39
have the square root again this is the same as saying the 1 half power isn't it so let me just change that okay again look at the square root right here we look at it as the 1/2 power we can bring the power to the front whenever we have a natural log but do not minus one this right here has nothing to do with derivative this right here is just a natural log property so all in all we will have to differentiate here is the natural log in the front and
3:23:10
then differentiate natural log of x and the 1/2 in the front and the natural log of 1 + x² like this in N algebra here comes to Here Comes the derivative 1 over natural log of x one over natural log of two sorry 1 over L and2 let's put down parentheses for the result of the differentiation derivative of natural log of x is 1 /x done for this part plus 1/2 it's a constant multiple just keep it and then differentiate this part so we have one over
3:23:55
the inside which is 1 + x² right inside but the CH says multiply by the derivative right here which is 2X and that's it we can cancel this out so that's pretty good but I think I combine the fractions I did so take a look right here this is pretty much it but I just combine the fractions and to do so of course we can just multiply this right here by this denominator 1 + x squ and of course do it on the top as well and for this one we need the X on the bottom as well so we multiply
3:24:30
the Bottom by X the top by X as well so all in all you will see the bottom how do I write the answer I put the natural log of two all the way at the yet right so I put natural log of two at the end I put down X first so I put down X and then 1+ x s and then here is the natural log of two okay and then we just have to work on the top one times this is just the inside this is just one + x² + x² so we have 1 + 2 x² 1 + 2 x squ that's it right 1 + 2
3:25:18
x square yeah I see the answer on the answer key so I'm good with my work and I'm also good with the answer key pretty good take a look take a look take a look
3:25:48
I feel like I write bigger when I'm doing derivative than um integrals I don't know if you guys find my videos helpful let me know thank you I appreciate that you can simply let me know by press the like button that would really help the video thank you so much though yeah and I'm having fun to do this as well and most importantly hopefully you guys do this on your own first and then check out the video that's the main part right okay number 5 five right
3:26:36
here differentiating here on the top we have x - one over x² - x + one three terms on the bottom but this is okay this is still a qu rule situation so go ahead do the eqution rule parentheses Square the denominator x² - x + 1 square Dot and then on the top man I have to I hope it's not so bad on the top we have to just keep the denominator and multiply by the function the derivative of the TP function and then minus the TP function which is x - one
3:27:25
time D derivative of the bottom the derivative x s is 2x der of- X is minus one the derivative of plus one is zero so that's what we have now let's see on the but bot we have this X2 - x + 1 s I think I'm still writing okay I don't know any anyway on the top you have this time St which is just x s - x + 1 but we have to F this out so let me just keep the minus in red right here and I'll open the parentheses x times this is x times this is uh just do this in
3:28:13
x * 2x is 2x s and then this is min - x and then this is min - 2x and this is + one okay and again let's see and we are going to finish this right now or do the computations in our head well actually I'll still write things down for you guys okay let's do it
3:28:44
x² - 2x² is x s okay next Negative X Plus X is zero but negative -2X is + 2x right lastly 1 minus 1 is zero so we're done so that's it I believe and yeah that's it for the answer and I left it like this and the answer choices as well so that's it ooh number 56 we have to in well no not integrate we have to differentiate one3 and we have cosine to the thir power X and uh minus cosine X like that well well keep in mind this right here is cosine X
3:29:51
to the third power right cosine X and then to the third power um so perhaps I will just make it clear for you guys right here we will look at this as cosine X to the third power right here okay so here we go differentiating this you bring the three to the front and then minus one three * 13 is gone just one actually so we have the inside which is cosine X but this time to the second power and multiply by the Der inside which is going to give us negative SX and we are done
3:30:28
for the first part okay and then here we have to minus the derivative of cosine X isga sin x h so that's what we have and now check this out here we have cosine and let's put on the Square right here and this is negative sin x like that right negative sin x and then we have the plus sin x like this H let's factor out the sin x to see what happens when we do that we have s x right this and that we Factor the out to the front so we have the s x all the way in the front well
3:31:11
well here we have negative cosine squar X plus this is out but don't forget we still have the one right here everybody knows well actually oh everybody knows 1 minus cosine squ X is equal to sin Square X but I know negative cosine Square x + one will give you sin Square X as well oh 1 - cosine squ so therefore sin x * sin Square X all in all we end up with s to the 3 power x and with that we are done but you guys don't see the equal sign so let me write
3:31:58
it down better for you guys okay you see the equal sign I'm happy here okay 56 no 57 O Okay e to DX * cosine X right here we go e to d x cosine X again you just keep it right just keep the E to whatever then you multiply by the derivative of of the inside pretty much is this part the exponent part multiply here we go power product rule s part product rule in action first function X multip by the D the second which is negative sin x multip by
3:32:46
negative sin x add the second function which is cosine x times the derivative first which is one I put on the answer before I tell you guys that's pretty much it what else did I do now that's pretty much it nothing too much we can do so this right here just know you just have to know how to write down the derivatives and this right here Negative X sin x and plus cosine X like this and with that we are done okay
3:33:25
58 this is done already number 50 no my God this sheet is done actually this sheet is done two more sheet but like this is the part I I'm I'm afraid of okay this is the part I'm afraid of to to be honest with you guys number 58 anyway DDX of X and we have minus square root of x * x + S root of x like that well do we need to use the product rule no because you can do the algebra you can multiply this out yourself first and yes check this out
3:34:12
this right here this is a minus B * A + B in that form right so the result is just a square minus b square namely we'll get x squ minus square root of x square which is just X so we just have to differentiate this instead in another word we can just say the answer is 2x minus one done okay next one 59 we are going to be differentiating inverse cotangent right inverse cent of 1 /x like this ah man I'm going to just see I have to think about how to save
3:34:54
space 59 differentiating inverse cotangent of 1 /x like that oh my God what's the derivative inverse cotangent don't worry it's just 1 over 1 plus x s right but you have to be careful of the input here we go derivative of inverse cotangent you get negative 1 over 1 plus the input being squared so here 1/ x and you square that it's very similar to the inverse tangent for the inverse cotangent
3:35:32
it's a negative right here but don't forget we have to multiply by the derivative of the inside derivative of 1 /x well it's like saying x to the ne 1 bring the negative 1 to the front you have negative 1 and 1 minus 1 is -2 for the exponent x to2 is the same as say over x square like that and in fact this also happens a lot you should remember it theity of one /x is this you can put that down right away that's pretty much it we'll simplify shall we negative * negative is positive
3:36:11
this right here 10 equal put on parentheses here we get 1 + this is 1/ X squar but you see this x square is going to be multiplying with that on the bottom so all in all you actually just end up with one over this times that is x s plus this times that is of course just plus one hey does this look familiar of course this is the same saying 1/ 1 + x² but didn't I say this is the derivative of something inverse tangent huh yes in fact this right here is the derivative of
3:37:00
inverse tangent of X well here's the question for you when the derivative of a function is equal to the derivative of another function does that mean that the functions are equal to each other in this particular case this is in fact the same as that but in many cases the functions might be off by a constant check out my other video for more detailed explanation right but in this particular case inverse tangent of X in fact you can write it as Cent inverse cotangent
3:37:38
of you can write this as inverse cotangent of 1 /x I'll prove this for you guys another time waiting for my 100 trick Identity or maybe trick uh prise that you have to know anyway number 60 so anyway number 60 okay okay okay okay I don't know if I can fit in here let's see this is a long one uh will put it down here maybe put it down here number 60 D DX X time inverse tangent of X and we are going to minus natural log square root of x² + 1 right so just like
3:38:35
that but don't we see a square root yes we do so I'm going to write this as the 1/2 power you guys know so this right here we are going to change that to the 1/2 power but this is inside of the natural log so can we put the one2 in the front yes right so all all I'm going to just rewrite this as Min - 12 natural log of x² + 1 okay so the other same now let's get to work well this part we need the product rule so keep the first function
3:39:15
times the derivative of the second derivative this is 1 over 1 + x² right and then we add the second function which is the inverse tangent of X oh sorry we keep the second function I should write down this part in Black we keep the second function times the derivative of the first which is just going to give us + one now for this part we have to minus this is the constant multiple so we can just write it down and then the derivative natural log of something is just
3:39:52
one over of whatever this is so x² + 1 but we are going to multiply by because of chend do the 2x right here okay so hopefully guys can still see this yeah that's pretty much it two two cancel now check this out all the way in the front we have X over 1 + x² next we have this guy which is inverse tangent of X lastly we have to minus X over x² + 1 what can we do yes cancel cancel and we just get inverse tangent for the answer very very very nice very very very nice of course
3:40:39
this does not not happen by accident I set this up the question is can you go backwards can you find the your what function will give you inverse tangent well you just did the r of this will give you inverse tangent that's in C to so don't worry so know this let's see if I can fit number 61 right here no I cannot so I will do the other one let's see so you will see why I not fit number 61 man after this video I will really have to think like twice or maybe three times to see
3:41:26
if I ever want to do 100 differential equations cuz that's just man I don't know
3:41:45
well shout out to Aran right AR again because he did 200 integrals in one take N9 and a half hours okay go check out his video I also have his uh the link to his video in the description he did it via a live stream it's I I don't know wow I'm speechless wow okay yeah I enjoy that video a lot because that video gave me that made me that me feel like how you guys are feeling right now like when you watch me like doing like 100
3:42:29
integrals 100 derivatives like yeah so shout out to Arian I pretty much pronounced your name wrong I'm so sorry Aran I think Arian yeah okay here we go number 61 differentiating and uh we have X squ < TK of 1 - x 2 / 2 and uh plus iners sign of x / 2 like that let me write down the question better x s < TK 1 - x 2/ 2 and plus inverse sign of X over 2 like this okay first thing first both of they have the over two do not use quion rule though because over two is
3:43:22
just a constant multiple perhaps it's better if we put the one over two all the way in the front you know that way it doesn't bother us too much so that we can look at x * square root of 1 minus x² and then plus inverse sign of X I think this right here looks slightly more slightly less interpretating yeah I don't want you guys to feel that oh man I just make the equ worse anyway here we go put down the one half of the way in the front open the parentheses
3:43:56
for the result of differentiation here is the deal we have to use the product rule keep the first function which is x times the thetive the second the of this is 1 over 2 root of 1 - x 2 but you guys all know the deal we have to multiply by the derivative of the inside which that will give us -2X it's 1 * -2X so this time I will just start to put down the -2X on the top save some space this way right so that's the first part and then we have to
3:44:33
add the second function which is < TK of 1 - x² and we multiply by the derivative of the first which is is just going to give us one like that and then this one is just going to give us 1 / sare < TK of 1 - x² that's the derivative of inverse sin x now the question is now what it seems that we have to get some common denominator again so we will get the common denominator isn't it so cancel this out first though right here let's just
3:45:12
multiply the bottom and the top by 1 - x squ also multiply the right here okay so we will see the answer right here is well the result is so far 1 over two times here we have X squ well everybody has a x everybody has a square root of 1 - x square for the denominator so let's put that down this guy has x² this guy we add and just put on let's just put on this guy different color this guy has Negative X squ first guy next guy plus this time St just 1 - x s right 1 - x
3:45:59
squ lastly we have to [Music] add this guy is just one so we have to add one no that's about all now well 1 12 well we can put the things stun together already so we will okay here's the equal sign okay here we go 2 * this is 2 * the square TK of 1 minus x² 1 + 1 is two this times that is -2 x square okay let's simplify a little bit of course we can Factor two and all that so reduce the two we get 1 - x 2 over sare < TK of 1 - x s right just reduce the two
3:46:49
just factor out the twos cancel out okay what can we do with this this and that have the same base this is to the first power over this is to the 1/2 power 1 - 2 1 - 1/2 is 1/2 positive though so you end up with square Ro TK of 1 - x s very good huh yeah in fact this is another calculus 2 equation can you find derivative function will give you one minus squ root 1 - x square as the answer okay I'll just do here 62 I want to have more space d DX here we have sin x minus cosine X
3:47:42
over sin X+ cosine X like this quo rule in action go ahead and square the denominator so we will Square the denominator sin x + cosine X and here is the square of the denominator okay the repon is kind of like tired okay so uh we'll see anyway on the top put this on the top so this right here we have parentheses sin x plus cosine X and then we have to multiply by the derivative of the top derivative of s is cosine X and the derivative of negative sinx is positive
3:48:41
sin negative cosine will give you positive sign okay so this is what we have then we minus the top function which is sin x minus cosine X then times the derivative of the bottom derivative this is cosine derivative this is negative sign so we have minus s right here wow Whoa man crazy looking uh insiration huh okay let's calm down and we shall be able to not read you uh I want to cancel things out let me just double check to see if I
3:49:32
happen of course you you cannot just say I want to cancel things out with so really making the top one okay I guess we really have to multiply things because on the top unfortunately we cannot Factor anything if you switch this it becomes positive that that's not cannot really factor I was thinking about to factor with difference of two squares but you know what let's just deal with this let's just multiply things out so so here we go s times cosine
3:50:15
right so sin x cosine X and then s so plus I know this is just s plus cosine Square I should have done that by too late and then this times this is plus cosine Square X and then lastly plus uh sin x * cosine X yeah so sin x cosine X okay okay now minus multiply this out we have S cosine this times that is minus sin s x and then uh I I know I should have just used the binomial theor I I seriously should have done that plus s cosine yeah this is sin
3:51:04
cosine can we see okay yes okay so this is what we have and all over the denominator I will just write it down like this okay I want to show you guys all the legitimate work okay now have fun to reduce simplify these guys this guy is just one Okay negative negative this right here is going to to give US1 all right oh check this out check this out check this out s x cosine X so that's two two of them cancel cancel ah so just imagine this will cancel with that this will cancel with
3:51:59
that in the end we're just doing one plus one on the top so Al together we have two on the top very nice over this guy for the denominator so I just put the downside more compact so that it does not look that intimidating I believe that's the answer and uh I do have that the answer choice that means I did it right I know the answer even though I didn't write it down I know what the answers is there's a nice pattern I don't know if you guys figure it out already you
3:52:32
can leave a comment down below and also leave a kind time stamp so I know you guys are watching this part number 63 differentiating 4X s times 2x3 power - 5x² okay ah it's a product but should we use product rule no because we can do the product in the algebra world first so let's go ahead just do the go this distribute distribute so 8 x s * x 3 power you add the poers so x to the 5th power this times that is minus 20 and you add the powers again so X to four now we can just
3:53:25
do the power rule right here nice D easy do not do the power rule right here and right there let me show you you see this is the right way to do it and you get 40 x to the 4th power this is -80 X to the 3 power with that you are done okay with this right here you are done do not right this right here do not just say hey this right here should give you 8X because that's the the root of this and then say hey this right here the the RO this is like 6X s minus 10 x
3:54:05
don't do this because you can see that when you multiply this and that this not that right this is not that this is not correct you can also do the product rule but the product root of course much harder it's not need it's not like super hard it's just there's no need for it okay so take a look
3:54:48
and as people often said differentiation is like a mechanical work right so it's not hard you can always just differentiate whatever you want but the good differentiation are the ones you have to actually come up with nicey like the ones actually actually simplified and has like some really compact answer some really nice answer all that anyway 64 D DX we have square root of s root x * 4 - x² we will do what we just did multiply multiply and then just use the power rule okay
3:55:28
so again you can look at this right here as x to the 1/2 power so let me just write down X to 1 12 power so we are just multiplying so we still have to keep the DDX X to 1/2 * that is just 4 x to the 12 and minus this * that 2 + 1 12 is 5 over two so we have x to the 5 over 2 like this now power rule power rule the go power rule right okay 12 * 4 is 2 and then this is X to the negative 12 here bring the one bring the 5 over2 in the front and minus one we get x to the positive 32
3:56:18
like this okay this should go down to the denominator so we will put it down in the denominator this is 2 over X to the2 power and this is - 5/ 2 and this right here is on the top so x to the 32 power can we get some common denominator right here and yeah I combine the fractions again so let's see this right here needs this two so multiply the two right here and right here this right here needs the X to 1/2 power so we multiply
3:56:51
this right here x to 1/2 power x to the 1/2 power okay 2 * 2 4 - 5 x to the 3 over2 * X to2 multiply you get x to the 4 over2 which is just 2 right for the power x square over 2 and X to 1/2 power is just square root of x just like that that's it okay number 65 differentiating square root of 1 + x over 1 - x like this okay here is the deal some people will tell you hey you can just code that to be Y and then take the natural log on both sides and
3:57:44
you can use the natural log property to reduce the right hand side you can do that that's not that bad but this is also not not that bad if you just do it so just do it like n key okay so here we go to do this 1/ two square root of the inside you literally just write it down 1 plus X over 1 - x and we have to multiply by the derivative inside which requires the quo rule multiply by let's Square the bottom so parentheses with 1 - x inside square that bottom
3:58:25
function here times the derivative of the top derivative of this right here is just one the inside function wise okay and then minus the top function which is 1 + x times the derivative of the bottom which is1 like that told you it's not so bad now check this out this is what I will do this right here it's the same as saying square otk of 1 + x over square otk of 1 - x I can bring this up up so we get square root of 1 - x bring this up up and like this and okay and then
3:59:07
divide it by two square root of 1 + x okay and right here that's see how we can simplify this here we have parentheses 1 - x s on the top here though this is just 1 - x and negative * negative makes it positive so we have to just add 1 add x negative X positive X cancel 1 + 1 is 2 cancel cancel finally we can put this together on the top we have square root of 1 - x over let's put this down first you can reduce more but oh man did I do that I didn't do it for the answer key did I do
4:00:03
that yeah I didn't do that but I could have done that but like I don't know this and now we can have reduced it more but up to you though anyway that's that's okay so one over there so I will have this and then one plus X I will reduce it and I'll change it on my answer key because I feel that's that's better here anyway this is like 1 minus x to the second power right but as I said we'll reduce this a little bit because they have the same base okay so this right here is the 1
4:00:43
half power and this is like 4 over 2 power okay so when you reduce of course you'll end up with three half power right here on the bottom so all in all you have one over this right here square root of 1 + x * square root you get three half power so you have 1 - x in the square root and you raise that to the third power right here okay so I will change that right here I'll change that on the answer key I just had this but uh I think I show do it like that okay same base just reduce
4:01:28
it all right number 66 this is a really fun one I will be able to uh fill in this right here number 66 differentiating s of sin x sometimes you can do a lot of them but you know two of them it's okay anyway here we go start from outside derivative sign is cosine okay derivative sign is cosine and the input you keep it you just keep it okay you keep it but the chle says you have to mply
4:02:08
by the derivative the inside derivative sin x is cosine and you have the X and if you look at the inside inside right here the derivative of x is just one so multiply by one doesn't really matter the truth is you are always using the Chand do but sometimes the CH do does not matter and uh on the answer choice I put down this first so cosine X time cosine of sin of X so we put sign in two cosine for this part and then this is just cosin X multiplying with st four hours okay number
4:02:53
67 okay the DX 1 + e to 2x over 1 - e to 2X
4:03:11
yeah okay of course quo rule in action so here we go Square the denominator before we forget so 1 minus E to 2x like this and square that and then we put down the bottom function right here which is 1 minus E to 2x right this is the bottom function times the derivative of the top derivative 1 is zero derivative e to 2x you write down e to 2X X first but we multiply by the derivative of 2x CH do so multiply by two then we just minus the top function
4:03:49
which is 1 + e to 2X and then multiply by the derivative of the bottom derivative of one is zero but thetive negative so this is still negative and you have e to 2x stays like this and then you multiply by the Dera 2x which is the two like that okay now let's see if we can simplify this nice D or not right take two e to 2x multiply in okay so multiply this into the parenthesis this times that we get 2 e2x and then this times that is -2 e and you add
4:04:35
the exponents so you have 4X for the exponent so that's that part then negative * negative is positive so be careful with it so we do this and also in the meantime we do this so negative together become positive so this is 2 e2x * positive one so this is plus right plus 2 e2d 2x 2 e to 2X and then positive times that it's also positive so you have two plus this is again plus two e to 2x * e to 2x is e to 4X just what over there and you have some nice cancellation
4:05:23
that's about to happen right here and here cancel them out and we can also combine things right this and that can be combined it so only all four e to the 2 X and you're adding so you keep the e to the 2x term you just do the coefficient the 2 * 2 the 2 plus 2 is four and then the bottom is parentheses 1 - e to 2X squ and we are done for that one okay I'm going to fit in 68 right here believe you or not I I I I I believe myself I I
4:06:09
hope DDX X over 1 + natural log of x another quion rule example okay I will do it right here otherwise I will be running out of space so Square the denominator so we have 1 plus natural log of x X and we square that and we put the button function up which is 1 + natural log of x multiply by the derivative of the top which is just one minus the top function which is x times the derivative of the bottom which is just zero and then plus 1 /x so it's just 1 /x like
4:06:58
this this right here is of course just minus one so you will see this right here is just 1 + natural log of x and then this is again as I said minus one over this guy ah so this is a very nice question because in the end 1 minus 1 is z we end up with Ln X over 1 + Ln X squ like that and with that we are done I I I have a question just now how can I integrate this and I get back to that that's a pretty nice question maybe do some use substitution this is a hard one I will
4:07:37
go back to this and then um do the derivative right here okay so this is uh number 69 I mean 68
4:08:10
all right here we go the next question all right this is a extreme fun one okay not because of the number but rather the actual question okay x rais to the X over natural log of x power like this let me tell you the answer to this right here is X to the X over natural log of x power and we are done
4:08:40
wow what is happening how can you have the derivative equals to the function itself there's only one function that can make this happen and yes this is e to the X this is how you do it look at the base remember we have X this is the function and this power is also a function ideally speaking we want to work with space e so we do this this is D DX this x right here let's write it as e to the natural log of x right and then we raise that to
4:09:16
the X over ah I know you see it now huh right because now when you multiply the powers the natural log of x will cancel so in the end you just have to differentiate e to the X power right the natural log of x will cancel which just just have to differentiate e to the X and of course e to the X will give you e to the X for the derivative and of course for this particular question let me just write it as x to the X
4:09:49
over One X to the X over natural log of x just for fun for that uh crazy looking expression okay hopefully nobody gets too M on this one number 70 uhuh let's see differentiating natural log and we have a square root so this is a natural log and the input is the square root and inside here we have x square -1/ x² + 1 like this all right so we do some algebra first to simplify yes we totally should so I will do the algebra right here DDX as you all
4:10:39
know this right here is the 1/2 power and now this time let's just do a lot of the algebra in our head okay one half power can be bring to the front and the natural log of a quotient can be split into a subtraction so first you will have 1/2 natural log of this input x² minus 1 and then 1/2 times the other one so we have minus 12 natural log of the bottom which is x² + 1 like this all right so let's just do the derivative now 1/2 is a constant multiple so just keep it
4:11:14
right here and the deriva of Ln of something it's just one over that thing right so 1/ x² minus 1 but we multiply by the deride which is 2x the CH do so and right here minus 12 again 1 over the Der this is x² + 1 in the denominator multiply by the derivative inside which is 2x like that and check this out this and that cancel this and that cancel and uh we just have this minus that so I will just write this down for you guys this right here is X over x² - 1 and- x over x² + 1
4:12:01
well if you want to combine the fractions which we will let's just multiply this one by X sare + 1 on the bottom and also on the top and multiply this one right here by X2 - 1 on the bottom and also on the top now let's see uh I will have the space right here let's see both of them have the same denominator this is a this is a minus B * A + B difference of two squares if you multiply out this and that we end up with X to the 4th power minus one all right and the topwise you
4:12:41
see this right here is X to a thir power let's put it down in Black that in red x² * x * x² is X 3 power m + x minus parenthesis this * that let's just distribute so it become easier minus x * that which is X cub and - x * -1 becomes plus X yeah so plus X in the end once uh X to the3 power minus X+ thir power can be canceled it x + x is 2x over x to the 4 power minus one we're done so it's a very nice one so it's kind of similar to number 65 but this one you do have
4:13:34
the natural log already so that's what me to be right ah 71 okay so this question un so bad it's just that I put like some second derivative and also implicit differentiation earlier now like oh man anyway inverse tangent of 2x + 3 here we go inverse tangent you get 1 over 1+ and you have to square the input right here right so put down 2x + 3 and we square that and what else do we have to do yes chendu multiply by the derivative of the inside multiply by this two okay so now
4:14:20
let's see what we get on the top we have two of course on the bottom this is the one plus let's Square this out binomial right so this is uh 2x s so it's 4X s right 2x you square that you also Al Square to two so it's 4 S 4X s and then the next term is plus 2 * this * that 2 * 2x is 4X * 3 is plus 12x and lastly you add 3 Square which is + 9 so it's just like that on the bottom 1 + 9 we can do that so on the top we have two over so again 1 + 9 so here we have 4 x² + 12 x +
4:15:07
10 let's put a two in the middle it looks better this way reduce reduce reduce we can Factor all the two and reduce it so you see if we factor out the two it becomes 2x s + 6x + 5 and then we can reduce this and that very nice in the end the answer is 1 over 2x² + 6 x + 5 and with that we are done
4:15:42
okay cotangent to the fource power uh of 2x number 72 uh I don't think we need too much space for this one so number 72 differentiating as I said it cotangent to the force Power of 2x right okay here we go um look at this right here as cotangent of 2x and then to the fourth power so perhaps I'll just write it down right here cent of 2x like this and then to the force Power okay here we go put the power to the front minus
4:16:24
one okay all right so we get four and the first part is cent of 2x to the thir power and then the CH do say multiply by the derivative this derivative cotangent is negative cant square and we keep the same input like that but the chendu of the chendu says we have to multiply by the D of 2x which is two multiply by two so it's technically like this you like this is a two because this whole thing is a derivative theide that's why I put
4:17:03
a r r parenthesis like that if you put it right here it's fine too it doesn't really matter that much anyway finally we have the answer 2 * 4 and then that's negative so we have8 this right here coent 3 power 2X and then cose squar of 2x that's it
4:17:41
ah complex fraction my God complex FR okay that's 73 D DX and hopefully you should do enough the riv equation so that you can get confidence like you can differentiate anything this right here we start off with a complex fraction but do Algebra first do algeb first so that our calculus life will be much easier so right here let's multiply the top and bottom by X so that we will see this right here is just DDX on the top is X Cub over x + 1 and
4:18:26
now this is just a nice qu rule question so Square the bottom let's put down parentheses with x + one in the Square that and then the top put down x + 1 times the derivative of the top which is going to give us 3 x² and then minus the top function times the derivative of the bottom derivative of the bottom is just one so that's what we have clean things up a little bit multiply this out we get 3x Cub plus 3x s and then - x Cub over parentheses x + 1 to the
4:19:05
2 power this and that can be combined so we will this is 2x Cub + 3x² hey just a reverse cool huh this right here is x + 1 to the 2 power and with that we are done not so bad not so bad not so bad three more sheets oh my God 20 something oh no man 28 questions I have to look this carefully because I just want to save paper so the print out is like tiny so this right here [Music] 70 uh see go under no I will just I just erase I will not be lazy I will erase I will
4:20:02
erase uhhuh it's like I Will Survive I Will Survive uh I don't know I remember uh I heard that music I will survive when I was running Mar I will survive I will survive I know even though it's not it's not meant to be like for like running what so but like it's it's just it fits that just say I will survive I will survive
4:20:39
number 74 right here differentiating DDX e is e x to the no X over 1 + x² power so for this one we just have to know do the outside and then like the quo Rule and all that so first e to the the dtive for that is just e to that whatever so e to the 1 over e to the X over 1 + x s then multiply by the derivative right here so multiply by and we have to use the quoti rule so we see that we have to square the denominator which is 1 + x² and square that multiply well down the top right
4:21:39
now we we we put on the F the bottom function 1 + x s and multiply by the derivative of the top derivative X is 1 so multiply by one minus the top function which is X and multiply by the derivative of the bottom derivative of 1 is zero derivative of x² is 2x so we just multiply by 2x so that's pretty much it now we just have to clean things up like we always do just like I will how to clean my hand later on so yeah anyway here we go this is just e to the X over 1 + x² and how do I
4:22:21
write that oh I put that all the way on the it's okay later later anyway put this like this right here so this is 1 + x² n square that this right here is 1 + x² and then minus this and that is 2x² here so here is the answer this is how wrot it I put this on the bottom only so 1 plus x² squ and I put this on the top so I actually had this first e to the X over 1 + x² okay and then of course multiply by one this is going to give us x² and with that we are done okay just like that
4:23:10
next one number 75 D DX here we have parentheses with iners sign of X inside and then rais to the third power please please please do not put down sign with a negative3 power right here okay if you ever write side3 power like this I don't know if you are saying this or are you trying to say sinx to the3 we don't know don't do this this is just so bad I know this is the the one that's the worst but just don't do it like that and for the regular calculus
4:23:52
course and whatsoever when you have this negative one right here it means the arc sign okay anyway here we go power rule first bring that to the front minus one so we have three parenthesis iners sign of X raised to the positive2 okay and then multiply by the thetive inside thetive iners sign is 1 / sare root of 1 - x s I will just put that right here just like that okay so with that we are done number 76 number 76 D DX I should start
4:24:43
training for my mer on to by the way haven't start running I have been going to the gym but haven't been running yet so my cardio needs to work right and this minus natural log of parentheses secant x like that very good so again whenever you have the power right here it's a good idea to write it as see X and then raise that to the second power anyway here we go bring the power to the front minus one 2 * 1 12 is just one so you have just this secant okay this secant x to the
4:25:25
first power so let me just emphasize this right here is from here to the first Power and then we multiply by the derivative of in the inside derivative secant x is secant x * tangent X that's the first part then we work on this part we minus we did the derivative this already we minus this part the derivative natural log of something is just one over of the inside which is secant x and again we multiply by the derivative the inside derivative secant x
4:26:01
is secant x tangent X well well what's Happening Here we can cancel the secant x that's nice and perhaps right here we can do the following this right here is of course secant Square X and we have the tangent X and then minus this right here is just tangent X so that's good and you see both of them have tangent X let's Factor it out so again this and that we are going to factor it out so that's in the front right here and then we will end up with secant Square X and then minus
4:26:40
one what's SEC Square x - one this right here is nicely equal to tangent Square X so all in all we have tangent to the first Power Times tangent to the second power so we will end up with Tangent to the 3 power x very very nice very very very nice as I said I set a lot of these questions so that they have some they have some really nice simplifications otherwise the derivatives they aren't so much fun next one this is a fun one it's actually a rather famous One natural log
4:27:25
of natural log of natural log of x it's just a lot of people would do this kind of derivative equation anyway here we go first natural log tells you you have to put down one over this is the inside you put it down in the denominator which is natural log and then natural log of x right and then the Chan says multiply by the derivative the inside this natural says you have to do one over the inside which is natural log X and then the CH says you have to multiply
4:28:02
by the Dera of this derivative of natural X tells you you have to put the X in the denominator like that was in the denominator like that so that's pretty much it in the end we just writeing down nicely one over this x goes first and then natural log of just X goes next and lastly natural log of natural log of x like that so this is it Dr Pam also did a video on this guy and this is the beginning of the chandu so if you guys haven't seen that video go check check that out ninja
4:28:38
milk number 78 we are going to be differentiating pi to the third power please pause the video and give this a try first okay good the answer is equal to 3 Pi squ easy right no the answer to this is just equal to zero why p is a constant when you raise a constant to a third power it's still a constant when you do DDX of Pi to third power you get zero for it that's it if you do that then you got tricked it
4:29:26
and usually this is a pretty tricky equation for the like AP not sure about AP test or like your AP class maybe your AP teacher might do this to you yeah I did this to my students too so I don't blame them all right number 79 I want to fit this right here okay oh my God the ones I'm scared they are coming the hyperbolic they are not so bad but the definition they are coming pretty soon actually for this one I'm going to erase
4:30:11
I'm going to erase I'm going to erase oh I will erase I will erase okay restroom break I'm going to just clean my head at this one one time like seriously it's it's just bad just all the markers and all that stuff so I will be back
4:31:35
for
4:33:33
feel much better I personally prefer chalk um 10 years ago nowadays of course I use markers but the downside markers is really dirty it's dirtier than chalks um if you have a bad one of course I mean like anyway okay number 79 here we go differentiating natural log of x + square root of 1 + x s I'm going to write this down better for you guys okay so here we go maybe I shouldn't be using my hand to erase the board that's
4:34:29
why it's like my was so dirty but anyway here we go this right here do we careful I mean it first of all natural L of something tells us to do one over whatever the inside is right so just put on x + square otk of 1 + x s so that's nice then we have to do the chandle now as you can see we have two terms but you guys cannot see because the marker is not so good anymore this to for mon if you're watching let me tell you I have a pink pen as well
4:35:09
check this out but I'm not using it I'm using a black pen of course black pen red pen of course anyway so it has two terms so I'm going to open the parenthesis hopefully it's better and the derivative X is one and we are going to add the thetive that it's not so clear
4:35:37
the derivative of square root is 1/ 2 square < TK of 1 + x s like this right however we have to multiply by the derivative of the inside for the square root of that part so in this case a common mistake is that people just close the parenthesis and they multiply by 2X and then they distribute no the Der of this is you just multiply right next to this part okay so this right here is for that derivative and after we have that close the parenthesis and we'll
4:36:11
see what we can do right here check this out two and two cancel that's nice and right here let's go ahead and combine the fractions so when we do that let me just multiply this guy by this which is square root of 1 + x² and of course we do the same on the top like that and after we have done that we will see this is 1/ x + S < TK of 1 + X2 okay I feel like my handwriting is getting worse
4:36:47
and worse I'm sorry 1 over x + square < TK of 1 + x² study better huh hopefully and we multiply this by check this out the bottom here they have square root of 1 + x² as the denominator so just put that down one time on the top this time that is just square otk of 1 + x² plus this is just an X right here right so of course you see that this part and this part this should be in the parentheses and the nice thing is that they are the same even though the water but you know the
4:37:27
order of addition does not matter therefore we can just cancel them out nicely and in the end we end up with the Answer 1 over square root of 1 minus so 1 + x squ this is it very nice isn't it maybe easy maybe you don't but it's okay number 80 this is the time now I will explain it to you guys yes this right here for number 80 from number 80 to number 90 it depends if you have covered the hyperbolic functions or not if you haven't you can skip this part
4:38:12
if you have you know watch this part of course anyway differentiating sign with an hge next to it this is not sign and H is a constant this is called the hyperbolic sign okay and the input is X like this and to make it more fun let's talk about the inverse hyperbolic sinx okay the inverse hyperbolic sign of X and again this right here is the input this right here is the name and we want the inverse of it so it's like that and I'll tell you guys more about what the
4:38:48
inverse in all the stuff is well the truth is this right here the answer is that because you notice this right here if you have the regular sign sorry let me just write it down right I'll I'll do it now WR down okay I have no idea what this is let me just say like that okay well how can I figure this out two ways first way I can do the following I can say that y equal to the inverse hyperbolic sinx like that okay and the reason I want to do this is because hopefully
4:39:28
we know the regular hyperbolic sign better well if so we can just take the regular hyperbolic sign function on both sides and we'll end up with this and then we can do implicit differentiation this is how you can figure out the derivative of infers functions okay when we do that here is the deal when we differentiate hyperbolic sign we get hyperbolic cosine so we have hyperbolic cosine Nam the Kos if you have a friend named Josh you know that's not cool enough
4:40:04
Kos is better and by the way this is s anyway this right here is why anyway I'm sorry about K and Josh Josh are cool as well but you know math math only okay anyway the derivative of hyperbolic sign is hyperbolic cosine named de just as cool as Co just just as cool as Josh there we go I don't know I was maybe if you are nearby and if your name is Josh by the if I See You by the if I still have this right I'll will offer you one
4:40:43
of this so it's see my office okay if you're watching my video if your name is Josh just bring your ID to my office I will offer you one of the Ninja milk okay anyway derivative of this is that but Y is a function of X so multiply by Dy d X isn't it and on the right hand side the derivative X is one implicit differentiation so from here the good thing is that you can see uh let me put it right here Dy DX is equal to of course we can divide C
4:41:25
of Y on both sides so we end up with one over C of Y but here is the trouble what is C of Y again if you are trying to differentiate this and this is in terms of X our answer should also be in terms of X as well so this is what we can do we are going to use identities so let me just put down some note for you guys okay for the hyperbolic trick functions they are just on the hyperbola
4:41:57
instead and C is the x value and S is the Y value and we start with x² minus y square equals 1 okay you can check them out video for more detail explanation by only all cos square of anything let me just put down t for the input also and then minus fin square of T this right here is equal to one so you can see that we can just put this onto the other side and we get cos squar T is equal to 1 + s Square t s Square t like this and of course you just take the square roots on both
4:42:44
sides so we end up as cos T and you want this to be square root of 1 plus s t like this this is an identity but here is the deal Technically when I take the square root on both sides I should put down a plus minus however the way that we define hyper body function is that we want K to be on the right hand side the hyperbola on the right hand side so if you look at x² - y^2 = 1 we just want the right hand side namely we just want X to be greater than or equal to one okay
4:43:21
and again C she is the x value since she is the Y value that's why you have this identity right away so in fact you don't need a minus just like that another thing you can see is C the exponential definition what the parametric way to define it is e to the t plus e to the T over 2 so in fact we are still dealing with like um exponential function know that they aren't so bad anyway so what um what am I talking about C of Y is just equal to let me just put it down here okay
4:44:06
we want the X to be there okay C of Y by this right here we just want the square root of one minus s oh s square there's a square right here s square of the same input which is y like that and of of course we still have the one on the top like this I'm running out the space man this is not let me let me try it again I'm sorry 1 over square root of 1 minus s square of the same input which is y like that well s of Y is what s
4:44:57
of Y is X therefore hold on this is a plus man sorry sorry sorry sorry C of Y is square root of 1 + cinch square of the same input okay and what's s of y s your Y is X so all in all we end up with 1/ square root of 1 + x squar okay so that's it the answer to this is just one over s otk of 1 + x s okay and the chuth is the chuth is okay uh this right here it's exactly the same as that okay it's not of by constant it's exactly the same and you can do that by doing some algebra
4:45:49
for so where you can check your out the video for it but anyway they are exactly the same so to differentiate this if you have this version for it you can just differentiate that which you did we did it already so this is it and remember the let me just say uh recall remember the regular version for the inverse sign if you have the inverse sign and this is the regular version not the hyperbolic version of X for the input right here you get 1 / square otk of 1 minus
4:46:25
X squ and there not hypo earlier it was myad that this is it okay you see this is the minus and here you have the plus when you have a hyperbolic version of it very cool very cool very cool so when I do the 100 trick identities perhaps I will just include some of the uh what's I called hyperbolic trick identities as well anyway number 80 that's the same as number 79 technicaly was SE number 81 for the rest I will just kind of show you guys the things that you have to
4:47:00
know along the way okay DDX e to the X Time s x and this is a regular version okay s of X well C is this s is just a minus in between and then the input is just X in Step because we're using X so to differentiate this right here we can look at we can you know Skip the product rule because this is e to the X and again for S of X this right here is nicely equal to e to x minus E to the X all over
4:47:37
2 so of course we can just distribute the e to the X first and do the algebra and then we can just be happy right okay when we do that let's put a one over two all the way in the front and then we have the differentiation we haven't differentiated yet e to x e to the x is e to the 2x e to the x * e to the X it's e to the zer so just minus one like that now for this one one2 all the way in the front differentiating e to 2x is e to 2x but the CH says we have to multiply multiply by
4:48:18
the D 2x which is two and the Der of minus one is zero so it doesn't matter two and2 cancel each other out so finally the answer to this is nice D equal to just e to the 2X very good huh 82 okay hyperbolic secant of 1/x oh my God hyperbolic secant okay of 1 /x that looks like a let me just write it down better hyperbolic I I don't want to have space there we go of x no no it's one of X okay like that well
4:49:19
well well the truth is I told you guys about K in s we'll be using them along the way this right here yes you are thinking it's secant H and then tangent H right kind of but you're missing the negative sign let me show you this right here yes you are thinking what exactly this is this is exactly the same as one over cos okay hyperbolic secant is the same as hyperbolic cosine one over hyperbolic cosine and then the input stats the same right here
4:49:56
of course then we can write this to the negative 1 power this is D DX of 1 /x then raised to the 1 power this means one over this does not mean the inverse C okay so again know your notations all that right here we can bring the negative one to the front minus one and go all that so we have negative parentheses COS of 1 over y 1 /x to the -2 power right here and then we have to multiply by the derivative the inside side derivative of cosine is negative sign but derivative of
4:50:40
C is positive s okay there's no typo the of C is s and you still keep the input and of course we still have to do the CH Lo right here derivative of 1 /x is - 1 over x² this is the one I told you guys you should know it by heart let me prove it to you guys real quick when you differentiate cos T you differentiate this e to the T which is still give you e to the T differentiating this you get Negative e to thega T over two is over two this right here is exactly
4:51:18
that which is exactly just s of X right just if you use a different input of course anyway this is the idea but of course we simplify this a little bit that's positive this goes down to the denominator this go stays on top so you have S of one / X this right here is a same as saying C of 1 /x right C of 1/x and another Co of one /x yeah and the reason I want to do that is because just like the Google days to prove the derivative of secant you see that right here
4:51:56
this is one over that which is by definition the hyperbolic of the secant and of the input which is one /x this right here yes is the hyperbolic tangent so you have the tangent and the input is 1/x and with that we are done very very nice oh I forgot the one /x oh I forgot x square I forgot this guy the x square I forgot this guy x square right here so we have to uh have the over this x square is still in the denominator and with that we are done for sure all right that's 82 now
4:52:42
83 let's see just for fun I want to fit in here 83 D DX of COS of natural log of x this is not so bad because we can just put a natural log of x in here and in here and we can take care of this check this out do the algebra DDX so putting natural log of x again in here and here so we end up with over two is still over two on the bottom so let's put that down and we have e to the natural log of x
4:53:22
so e to the natural log of x and then for the other part for the other part we have the Plus
4:53:35
e to the negative okay natural log of x like this well well we can simplify this and just have to do it carefully because for example e to the natural log of x they cancel nicely but this and that it does it does not cancel nicely yet you have to put the negative right here so you end up with e to the natural log of x to the1 power and then cancel you out you have to make sure this is right here is clean it has to be just a positive one technically so
4:54:10
bring that to the front so you have that so we see this right here it's actually just to have the one over two all the way in the front and then DDX first part is we just have to differentiate X second part we have to add it with X to the1 power so this is is all we are differentiating so here we go 1 12 all the way in the front put down the parentheses for the result of the derivative derivative of x with respect to X is one der of this is -1/
4:54:48
x² and you can just get a common denominator real quick in the end we end up let's see 2 x² on the bottom right and then this is X X sare - 1 * 1 still x² - 1 like that whoa look at that I told you this should be able to fit in there let just take a look take a look take a [Music] look okay let me know if you guys like the hyperbolic trick function more what the regular trick function
4:55:37
and also let me know what kind of derivative I should be doing please do not ask me to do like x to the X to the X to the X to the X I did them already actually I actually should have done that but like yeah uh yes I will be doing 100 limits and I have a lot more ideas but it also takes a lot of preparation and energy and also time and uh yeah so uh I'll try to to see if I can do them soon as possible
4:56:06
number 84 I kid you not I switch the natural log and also the C you guys see okay yeah so 84 D DX of natural log of C of just X all right go C okay here we go go natural L however this one we cannot do the algebra in the sense that do not write it as e to X um plus whatever whatever hold that doesn't really help check this out because we can actually just answer the answer the question in the hyperbolic trick function World thetive natural of something is just one over whatever
4:56:56
that is so just write that down then take the derivative of the multiply by the C is s and it's positive so just multiply by s x guess what you're done because actually you're not done yet you can write this as s of C which is T what hyperbolic tangent of X that's it that's it that's it that's it that's it the identities are similar but they're not exactly the same so be careful with that number eight 85 I'm not exactly I'm not like super happy because I know
4:57:39
the hard ones are coming so yeah anyway s man we have to make sure the H is not too far apart from the S otherwise it looks bad one plus we did this version question number two okay long time ago uh but this is the hyperbolic version of the question here we go horrible sorry okay here we go Square the denominator so put a parentheses and then with one plus with 1+ cos x inside right square that and now on the top you of course put on the
4:58:30
denominator 1+ cos x times the derivative of the top derivative of s x is just cos x and then minus the top which is s x times the derivative of the bottom the derivative of the c x is positive s x so you just multiply by s x like this can we expect some kind of cancellations well watch to find out here we continue this times this this times that COS of X this times that of course we write it as cos square right just like the trick function you put a square right here of X and
4:59:16
then minus this right here is s Square X all over this right here parenthesis 1 + cos x and square that yes very good this right here as I mentioned earlier it's nicely equal to one and we just have one of one and that right so you see I can put down the one right here first and this feels like dju but it's not exactly anyway this is equal to okay this is equal sign you can see now over this guy parentheses 1 + cos x squ cancel this cancel one of that so finally we get one over
5:00:05
finally we end up with just 1 over 1 + cos x right here and we are done ah number 86 this is one of the my my favorite one so we are going to differentiate okay derivative of secant is secant x time tent X but can you give me derivative what function will give you see X in fact we saw it earlier but we're going to see again right here check this out differentiating inverse tangent sorry inverse hyperbolic tangent and this is the inverse hyperbolic tangent of cosine
5:01:05
okay uh in fact I said it wrong derivative of cosecant is equal to negative cosecant time cotangent but can you can you tell me thetive what will give us cosecant or maybe negative version of it watch to find out anyway anyway anyway anyway anyway so this right here I will just tell you guys what is what the deal is okay in order to do this okay you will have to know the following so here's some notes on the side DDX of the invers tangent and this is a regular version okay this
5:01:42
is 1/ 1 + x s well take a guess of what the derivative of the inverse hyperbolic tangent of axis no it's not a negative sign right here that's inverse cotangent that will do the work this right here get 1/ 1 - x² like that very very nice so same idea but you know just do the work right here same idea same idea okay so to differentiate inverse hyperbolic tangent we have to have one over 1 minus we will have to square this guy so I will put a parenthesis with
5:02:24
cosine X and square do this is cosine X okay and we multiply by the derivative of the inside which is negative sinx now check this out this is just one over this is 1 minus cosine squ X and perhaps I can put a negative sin x on the top like this right but what's this we saw earlier this right here is just nice to equal to sin Square X right this is just sin Square X 1 - cosine square is 1 1 - cosine square is sin Square what's on the top sinx so can we just cancel cancel yeah this
5:03:16
is going to be -1 over sin x in another word we get Negative cosecant X so in fact if you want to find out derivative what will give you cosecant or secant you can use use the inverse hyperbolic tangent to help you out right you can just have think about what other things you have to do with the with this so you can end out with uh the derivative being secant or cosecant and maybe a negative version just multiplied by negative anyway that's the answer for that 87 87
5:03:53
87 man D DX I haven't done any product rule and all that for the hyperbolic huh don't worry I am going to do it x * inverse hyperbolic tangent of X and plus natural log of square Ro TK of 1 - x s like this well you guys all know the deal this right here is the 1 half power so I put that all the way not all the way please right here right so I will just change that for you guys one have
5:04:33
natural log of this guy and now let's get to work here is the product right here x times the the to the second it's right here so we multiply by 1 over 1 - x s Chu says you have to multiply by one but you know you don't have to you one multiply by one doesn't matter and then you add second function which is inverse hyperboy tangent of X and you multiply by the derivative the first which is just one so that's what we have now we add the Dera of this part this is
5:05:13
just2 times the natural will give you one over and you put the inside in the denominator and the CH says multiply by the derivative right here which is -2X just like that this and that again cancel out you guys will see why I say again this right here x over 1 - x 2 this right here plus inverse hyperbolic tangent of X this right here is negative though negative okay so - x over 1 - x^ 2 this is positive this is negative of course they can say bye-bye to each other finally we
5:06:01
end up with the inverse hyperbolic tangent of X and this right here is your nice answer very very nice similar looking than the regular inverse uh tangent X we did that earlier as well okay I am going to do the next two side by side which one you guys are see this right here right I'm going to do the next two side by side so I will how to erase the board and again I'm not so happy because but I'm kind of happy because it's almost
5:06:42
done but like the hard ones are coming so I'm not exactly like super super happy this is like I don't know whoever decides the course they put like uphill at the finish line and that that's that's really mean okay I would rather have the uphill um then the first stre mile was so and just just have it done with it yeah if you run you know what I'm talking about anyway number 88 I'll put it down right here okay here we go D DX we are going to be differentiating inverse Sy okay
5:07:26
the first one right here is inverse s and the input is tangent X okay I am going to just treat myself better I will have to I would like to have a better marker right here so I can feel the writing experience is better so I'll use this one ah much better okay so number 88 D DX inverse Sy of reg tangent X like this yeah hyperbolic together with the regular so you're are going to see what's going to happen and on the contrary on the other hand I'll say
5:08:14
number 89 we will [Music] differentiate the regular sign of hyperbolic tangent x o what in the world is happening huh so again I'm just looking at the questions I just want to make sure that I wrote down the correct question now start with this one here we go inverse cinch is one over square root of one plus this plus okay and then you put this right here and you square that so tangent square and then what else do we do or we write write down the tangent X
5:09:06
better and what else do we do we multiply by the derivative of the inside CH say so derivative of tangent X this is a regular tangent X we get regular secant Square X okay so it's just like that okay this is good but see anything that we can simplify yes because 1 + tangent squar this right here is nicely equal to secant Square as well so you are looking at the square root of that which is just a secant x in the denominator all in all you just have
5:09:51
secant squ x over secant to the 1 power x right after you take the square root and also the the square they cancel and of course we just try not to uh deal with the absolute value don't worry about it whenever this things happen for now you have want to be more technical you have to look at the actual domain and also the X values that you are talking about for this one just just just cancel out yeah anyway this and that cancel one of them out and you end up
5:10:22
with what secant x thetive of this will actually give you secant x we have saw we have seen the the some function will give us c x earlier as well like I don't know how many hours ago but yeah anyway let's take a look of number 89 take a guess what the answer is okay we'll see okay this is the regular inverse sign so we get 1/ square < TK 1 minus this right here we have the parenthesis with Tangent with h so it's the hyperbolic tangent okay okay X and uh we square
5:11:04
that and we multiply by the derivative the inside derivative of the hyperbolic tangent yes is the hyperbolic SEC Square X that is true you can prove that with uh whichever wether you want but what is this yes identity yes this right right here it's also the same as hyperbolic secant of X but you have to make sure it's a minus 1 minus uh T square is actually inverse hyperbolic uh is a hyperbolic secant x square right square and just like that you take the square root so you're
5:11:56
looking at on the top hyperbolic secant Square X over on the bottom is just to the first power so see hyperbolic secant of X cancel cancel this is one so you have hyperbolic secant of X so cool huh you just put H somewhere and you can get the H somewhere else too like that very very very nice okay the last multiple choice question uh let's see I would put it down right here ni 9 D DX here
5:12:36
we have the inverse hyperbolic tangent of X over 1 - x² like that okay don't be too excited we are not doing integral you don't do your substitution for this but we'll do quo rule for this okay here we go right here we Square the denominator here we Square the denominator on the top we put 1 - x² the denominator right here and then multiply by the derivative of the top function which is 1 1/ 1
5:13:11
- x^2 like that and then we minus the top function which is inverse hyperbolic tangent of x times the derivative of the bottom which is going to give us -2X okay and again this x is the only input for the inverse hyper body tangent well well this and that cancel nicely so we don't have to worry about the complex prodction situation we just have a one right here negative * negative okay so plus 2X and
5:13:47
this is the hyperbolic inverse hyperbolic tangent of X and then on the bottom here we have this guy which is parentheses 1 - x² squ and with that we are done we are done with all the multiple choice questions I'm not so happy as I said even though I'm almost done but like I know the hard part is coming so I will have to calm down I will have to make sure I can get ready for the next 10 question I don't know how long it will
5:14:27
take it's 5 hours only it's not so bad 5 hours and 40 minutes I just saw the timer but the good news is last page okay ladies and gentlemen welome to take the derivative with definitions so number 80 sorry number 91 right so we are going to be differentiating DDX again and x to the 3 power I know the answer you know the answer everybody knows the answer but do you remember how to use the definition for the answer no you don't then watch out okay and you
5:15:13
just have to make sure you follow direction see a lot of students they lose points for this kind of questions on the exams because they don't follow direction if the question say use the definition you have to use the definition otherwise you just you're not going to get any points okay seriously don't be q and say the answer is 3x to the second power no you have to show the work on the exams so
5:15:38
you have to remember you have to remember okay you have to have to know that and why do you need to know this well that's how you can prove things with definitions anyway here we go this is the function this is the input the x is the input well you take the limit as H goes to zero and remember the difference equ is f ofx + H minus F ofx all over H so you put x + H into X right here so you
5:16:11
open the parenthesis with x + H here and you raise that to the third power and then you minus this right here is just that so it's x to the 3 power and then you divide it by H like that okay now you can just multiply this out you can also try to factor it but I think multiply it's easier for for this situation you can actually use the difference of two cubes to factor but up to you anyway you
5:16:42
have to write down the limit as H approaching zero and when you multiply this out you just do it somewhere I'll show you just to make sure that we know what we are doing okay X plus h x plus h X Plus H so for the first two part of course you just multiply this out so you end out with X squ and then you get xh plus another xh so you have two of them two xh and lastly you have H squ all
5:17:19
right and you have to multiply the one more time so X Plus H so you do this time that which is okay let's do this time that which is X Square * X which is x^ 3 power and x squ * H so it's plus x² H and then this times that is plus 2 x² H and this times that is plus 2 xh s I know the passal triangle I know the fast way but I just want to show you guys all the work because I know not all
5:17:59
the students know about this okay so that's the this times that is x * h² and then this time that is plus h to the 3 power like that all in all we can collect like terms and put the result right here so when we do that we'll see this right here is X to the 3 power and let's see this is x² H and we have two more these two things can be combined so on all we have two of them so we plug
5:18:31
plus 3 x² H and we see that this and that are like terms 2 + 1 is against three so we put on+ three x h² and lastly we have the plus h squ not H to the thir power right here this is the first part right and then we have to just minus x to the 3 power and now we're just literally minus the X to thir power that's good huh and as you can see everybody on the top has an H so you can Factor it
5:19:10
out and good things are about to happen so check this out here we have the limit as H approaching zero f out the H and we have 3 x² and then F out the H we have plus 3 xh and then fact of the H we have plus h squ and then this is all over H how can we do with the hes cancel them out cancel this out lastly remember X okay remember X is the variable remember that we are taking a limit as H
5:19:43
going to zero so we just put zero into all the H the remaining H and you see when you cancel this H and that H there's no more zero over zero so no nothing okay no worry okay so you have 3x² and plus 3x times putting zero in here and lastly you put down plus 0 squar this is z that's 0 yes we get 3x² okay and we are done this is actually not that bad I know yeah I wasn't going to do this in
5:20:22
the first place but I know I I am sure some of you guys are going to say Hey where's the definition of derivative yeah so I feel like I have to so anyway this right here is 3x² let me see if I can fit in one more right here number 92 I'll give you gu another type of the question this is another one that we did um in the first couple question I'm not going to bother to look it up because um just going to finish this and I want to go home anyway D DX square root of 3x + one here
5:21:03
is the input which is X right here so you are going to plug in X plus h in there and then do the rest anyway take the limit as H going to zero put in X+ H right here so we have the square Ro T 3 parentheses x + H and then + 1 and then minus the function which is just square root of 3 x + 1 and everybody is divid by H right here right so uh my H is okay okay yeah let just want to make
5:21:42
sure I'm okay now we are going to just see how to simplify this kind of limit to do so when you have square roots keep it in mind if you plug in 0 into H right now you get z z one way to simplify this situation when we have square root is to multiply the top and bottom by the conjugate so that's the usual strategy so we will do so so multiply the top and bottom by the conjugate which is
5:22:12
just this plus that square root of three x + H + 1 and then plus Square < TK of 3x + 1 and of course do the same thing on the bottom so parentheses square root of 3x + h + 1 plus square root of I like to keep my color consistent and then the 3x + one like that okay here is the good news to multiply out the top it's not so bad because you are in the form of a minus B * a plus b so this is going to give us the following we still have the limit as H approaching zero well when well on
5:23:02
the top we just have to do a square minus b square a square is just the inside well in the meantime let's also distribute that power that actually let me just keep it how it is three parentheses x plus h plus one okay a square and then minus b square B Square which is just this thing square and that's just going to give us the inside and let's put a parenthesis because I will have to distribute the negative in a second that's what we have right and for the bottom just go ahead
5:23:36
write it down again because remember the point of multiplying this conjugate is to fix the top so just multiply out the top and for the bottom just leave it like this and wait for good thing to happen so you will see the good thing will happen this right now let's just multiply it we get 3x plus h 3 H and then plus one and then Plus square root of 3x + 1 it actually doesn't matter if you multiply that or not but whatever anyway right here here we go this right here is 3x + 3
5:24:17
H + 1 - 3x - 1 yeah can we reduce yes you rally cancel the 3x and also the the 1 minus one is done yeah and now check this out on the top we have just three h left over H * this quantity this is three * H so we can just divide the H out so we divide this out Okay cool so no more zero over zero situation we just have that three on the top and now we can just put a zero into the remaining H which is only this one here keep in mind we still have this three right here
5:25:01
though okay we still have this three right here this three this three right here three over well I putting Zer into H so we have square otk of 3x + 3 * 0 + 1 and then plus Square < TK of 3x + 1 like this don't forget this three this is how you can show that you need a chain rule need a three in the end of course this is right here just zero and you have 3x + 1 inside here in the square root
5:25:40
and they are the same square root so we can have 3 over 2 square root of 3 x + 1 and we are done okay that's why you see we need a chain rule we need to multiply by the derivative of the inside 3x + 1 which is going to be the three right here on the top okay chain rule matters number 93 I need more space uh no not really I'll see I will try to fit in 93 it's a seriously a lot of fun to just try to fit in a lot of like crazy computations and Aller things 2x + 5 in the denominator
5:26:26
anyway here we go limit as H approaching zero here is the X so I will put down X plus h in here so we have one over two parentheses x + H and then + 5 and then minus 1/ 2 x + 5 and uh all over like this now take the limit as H approaching zero I cannot regret but uh I will try to just deal with this we are going to multiply the top and bottom by the LCD okay so yeah I don't know if I have enough space okay we have done a lot of LCD already so let's not show work for
5:27:27
this one okay so I will just show you guys what we can do
5:27:33
here is the deal h on the bottom H you multiply the top and bottom by this and that namely dlcd and when you do that you pretty much have parenthesis with this Nam the two parentheses x + H + 5 that's this denominator and then you multiply by 2x + five yeah and then remember we are multiplying this this and that on the top and bottom when you have this multiply with the top you have this right here only because this and that will cancel just multiply out you know you
5:28:11
pretty much have 2X + 5 right here and then minus you pretty much of this right so just do this and do that that's the fast way to do it minus parentheses and here you have two parentheses x + H why why don't I distribute I don't know man anyway this is what we have now let's just clean things up this right here is two see this is why why don't I distribute I don't know okay I'm not higher I can still do this all right so here we go 2x + 5 and minus you see this is 2x
5:28:57
plus 2 H but it's minus so because- 2x- H yeah okay and then minus 5 okay and you see 2x -2X cancel 5 andg five cancel you have the -2 H this H and that H hey hey we can cancel the out again just like over there so be sure you cancel out this right here and you cancel out this right here it's the ID sign you know three cancel like three cancellations and you get the negative two on the top don't forget about that now on the top we have -2 left on the bottom we
5:29:37
putting zero for the h right here that's about all so we get parentheses 2 x +0 m m + 5 and this right here is another 2x + 5 of course if you distribute it's the same thing at the end this right here is just 2x + 5 and and 2x + 5 on all we get -2 over parentheses 2x + 5 to the second power now you can see that this kind of quo rule or the if you bring this to the negative one power all that and also the chain rule they all work legitimately all right don't be afraid of using
5:30:24
the definition of derivative this is like all the calculus teacher the things that we want to tell the students okay for the next one I will have to erase the board I put on this is too long so yeah
5:31:09
1/ x to the 2 power right uh number 94 again I'm not so happy because the harder ones are coming but it's okay we can survive here we go X is right here so we are going to write it down I put it down right here why not I think say some more space the limit as H approaching zero and we will just have one over parentheses x + H square that guy - 1 / x² over H right here and that's practice what we did over there okay so this right here you
5:31:52
multiply the top and botton by this and that namely the this common denominator limit as H approaching zero well well you put this right here put this right there so we get x squ minus you put this right here so it's parentheses x + H squ okay and then the bottom is just H in Black times this and that so I will put down x square first parenthesis and we have x + H squ like that
5:32:31
okay here it's a common mistake some students they just like to cancel this out and then cancel this out and no you are forgetting the minus sign in between so don't do that don't do that you can do the difference of two squares you can factor out and you can simplify it that's okay but I find it easier to follow in the sense that if you just multiply this out and then just reduce so that's what we will do so let me just write it down again this is the limit as H
5:33:01
approaching zero and we have X squ and this is minus when you multiply this out this is just going to give you x² + 2 xh + h² and we have all this Nam the H * x² * x + H squ like that now be really careful you have to distribute this so it's actually x² and then -2X h and netive h² well well this is x square this is negative x square so they can be canel it on the top they have the H left so we can Factor an H so let's see this is the limit as H approaching
5:33:50
zero when we factor out the H we will have -2 x minus H2 to the first Power like this over this H * x² * x + h² like that so good the moment cancel out the H with the H so good now we can just put a zero into the rest of the H and we'll be done and just simplify if you need to on the top is -2X minus Z on the bottom this is x squ * X plus zero in here and we square that well well on the top is -2X on the bottom this is X squ this is also X squ multiply them we get x
5:34:42
to the 4th power this and that can be canceled it finally we get the answer -2 over x to the 3 power I know what you guys want me to do I know I really do DDX hey this is just X to the2 power isn't it why don't I just do this front minus one and we get -2 x to3 which is the same as that I know I know yeah let's drink some Ninja milk and then we will all know finway use the definition use the definition number okay now these are the questions I'm like like I I didn't put it
5:35:42
any easy one at the end uh I will just show you s x okay here is the input x s of X they are not so bad it's just a lot to write especially once it's like 90 something question it's just tiring anyway take the limit as H approaching zero on the top we have sign open the parentheses here we have x + H minus s of X like this this time be really careful we are using H as a variable it's not Sy okay not yet no we don't do we don't do definition with a S no no no no no just just
5:36:27
just just regular sign okay anyway this is over here is the deal do not multiply the conjugate do not get whatever the way we do is look at this right here we have the angle Su formula for sign so we'll do that with that that this right here it's a limit as H approaching zero let me just tell you guys that sign of a plus b is sin a cosine B plus sin B cos a so here s of X Plus H let me write it down we have sin of X time cosine of H and for this one I really have to
5:37:10
put a parenthesis just to make sure that you guys don't this is not Kos anymore Kos left us already Okay this not Co it's cosine of H cosine of H and then plus s of H okay s of H times cosine of the first which is X like this this is the anle sum formula for sign the regular sign and then minus s of X like this all over H right here so this is all good now here is the de if you take a look H let's see this one has sinx this one has the minus sinx as well this is what we like to do
5:38:04
I'm going to pair this up with this okay and uh you see the following when you are talking about H is approaching zero that means what in the H world in the limit sense okay that means X it's like you can just bring to the front it's not affected by the H so I would write to write this down for you guys perhaps I said it's not hard it's just a lot to write
5:38:32
let me write it down this is the limit as H approaching zero I'm going to put this and that together but because both of them have the sign so I can factor out the sign of X isn't it okay and right here we have cosine of H cosine of H minus one after we factor that out and then we have this over h of course because the denominator is over H and then we can say we plus I'm just splitting
5:39:07
the fraction right now here we have sin H cosine X but let me write down the cosine X right here let me just write it down like this sin H cosine X and this is also over H and we are taking limit for this right here right now by the limit losss we can say the limit of the sum is the sum of the limits if both limits exist and also the limit of the finite product is the finite product of
5:39:40
the limits if the limits exist okay so I did say that and now I'm going to write it down in action for you guys what I mean by that is I can say the limit of this plus the limit of that okay and again H approaching zero the X doesn't matter so I can put the sinx in the front first s of X right here and then I'll just focus on taking the limit as H approaching zero and we are looking at this part namely cosine of hus one over H okay and then we are going to add we are taking a
5:40:19
limit of this guy okay we're taking a limit of this guy as H approaching zero but notice here we have the cosine of x in the H world the coine X doesn't matter so we can just put that in the front right we we're not looking at the there's no h right here and then the left over right here is s of H over H like that all right now this is the part that has a lot of the controvers this is the most controversial part what's the limit of this what's the limit of that especially this one
5:40:58
I coded the limit you don't do lito Ru to figure out this limit you don't do like derivative Taylor series whatever to do this because we're trying to figure out the derivative sin x if you use Lao rule you have to know the derivative s x in the first place but we don't so let me just tell you guys uh in blue this limit right here okay this limit right here is equal to one okay and and this limit right here is equal to zero one way you can do it is you can just let for example
5:41:39
H equals 0.0000001 and put it into here and put it into here and you'll see the number is approaching one that's the limit the limit is one of course you have to just do even more zeros and you'll see it's approaching one I will be much happier if you do this kind of things right here and say the answer is one rather than hey lapos Rule and all that stuff no and keep in mind H right here has to
5:42:05
be in radians why I will let you guys know later on I'll do another video for it so this is one way to do it in the you know just use a calculator for it quickly or you can also do geometry okay or secondly I just put or do some geometry geometry okay and I do have a video on this right here and I do have I actually have a t-shirt with the geometric picture the picture that you need I will have the link in the description for you guys the triangles the sector and the connection
5:42:41
between s Theta Theta and tangent Theta all that stuff and squeeze theorem do geometry and squeeze theorem okay so hopefully you guys okay with this little explanation and then I will just continue to figure out my derivatives all right all right finally you see that s of x * Z this is all gone Z 1 * cosine X finally this right here is just cosine X ladies and gentlemen the r of the sinx is cosine X I know I know I know and then what I'm going to do next is I'll show you guys
5:43:24
how to prove the derivative of secant x with the definition take a look and then remember this two limits I'm not going to give you guys too many explanation anymore because I did that already he man that's the part that's like it's hard to convince people in the uh this called circular reasoning so it's kind of weird it's it's it's yeah okay Number 96 secant x okay number differentiating because I've been using derivative secant x a lot
5:44:11
I don't know if I have been doing it right or not maybe I should go back to redo the whole thing if the the secant AIS happens to be like I don't know tent to the fifth power I don't know we'll see anyway here we go the ver
5:44:31
here we go the definition of derivative we take the limit as H approaching zero and as you can see this right here is our uh input so we have secant of x + H and then minus secant of X or over H right well we don't really have a nice formula for the angle sum for secant because we should look at secant as one/ cosine so we'll do that that's the deal so here this is the limit as H approaching Z this right here is 1 over cosine of x + H minus this is of course 1 over cosine X
5:45:17
and this is why I didn't do cosine because we are seeing cosine right here as well now let's do the
5:45:31
following I'm just going to multiply the common denominator and I'll tell you guys how to expand this in the meantime too okay so multiply the common denominator so this time I will really just put this down for you cosine of x plus h * cosine of x and here cosine of x + H cosine of x right here we have the limit as h approaching zero man my limit is like getting worse sorry the limit
5:46:04
as H approaching zero this times that of course we just have the cosine X on the top and that's why I told you guys earlier for the uh complex fractions situations now when we have this times that of course the coine X cancel we have minus and let me just tell you for this part right here we can expand it and the way to do it is we get cosine of the first angle which is cosine of x time cosine
5:46:34
of H cosine of the second angle and then if it's plus it's weird right if this right here is Plus in fact here it's minus you have to just remember that okay or you can see the proof for it you'll have video for that as well and then secondly you do sign of the first and the sign of the second yeah so just like this okay so all over this is the H all the way in the front and we have this part again so you just write it down again man it's a lot I know cosine x cosine H minus sin x s
5:47:16
h so that's that part and then of course we still have the cosine X right here right so hopefully you guys are okay with all this and maybe I'll just might as well change the cosine X in blue so this is just some of the prep work for that now we have to just kind of play around with all this yeah here is the deal um we are going to split things up right and you see when H is approaching zero in fact you still end up with zero over zero situation if you're just
5:48:08
plugging zero into all the H and again just earlier we can take out the parts that has nothing to do with the um what's that the X with with the H and all that stuff okay so hm
5:48:40
how should I split this in the nicer way let's [Music] see okay man as I told you this is just a lot to right so unfortunately this has to stay inside of the limit so it's that but this can go out but here is how I will split this okay so multiply this in and uh check this out I have the cosine x cosine X so we can factor that out and then we can Pier things up the hard part is to Pier things up the right weight um okay this is like the
5:49:21
limit law to the finest the limit of a product a finite amount of product is the finite am am the product of the limit so here's the deal I'm not going to touch the H I'm going to split this in fact if you have H equal to approaching zero right here this is just cosine X likewise this is just going to be zero so this is good so this is one part that I can do so I would like to just take the limit of this part I will show you limit as H approaching zero and let me just forget
5:49:56
about the tough now okay and also the H just not do that for now so I would like to look at this one over this whole thing here this whole thing here man perhaps I should have just written like that huh oh man let's see if I should do that I should totally do that man I should totally do that ah so mad at myself H I am going to just keep this right here you'll see why so H cosine of x + H and this is cosine of x so what I will do is
5:50:38
I'm just going to put down this right here first okay so we have one we have one over I just want to put this down right here okay so this is cosine of x + H cosine of x so that's the first part I want to just keep it as how it is and I will have to multiply by the following the limit as H approaching zero and we will see the rest so you see I am putting down this part right here so this right here is just the one that with h in the denominator I put on
5:51:19
this two in the front already and that's going to be nice now I will have to just pair the uh the top on top I just have to pair the top nicely on the top here cosine x minus cosine x cosine H right I will just Factor the cosine X right here so I will put down cosine X and this is the cosine x cosine X and then parentheses this will be a one minus cosine H good huh yeah of course it is good
5:51:59
then this is negative and this is negative so I should have done it like this negative * negative becomes plus so we have a let let me just write it down again like plus and you have the sin x right here time sin H and I think you guys know where I am going yeah that's pretty much the idea right so for this part in fact as I said if you can figure out limit go ahead and do it and that's pretty
5:52:36
much the limit law putting zero into the H we will end up 1 over cosine x + 0 cosine of x and that's it right this is the first part of a limit let's just ignore that for now and now if you look at this part of course course let's just do it like this okay save some space save some time just like what we did earlier the limit as H approaching zero this is going to give you zero so this is
5:53:09
nothing here so multiply by zero and this is going to give you one so we end up with just sin x right here so it's plus sin x right s h over H the limit of that will be one so it's just s x remaining so it's like this now we see that all in all we just have SX on the top over this is of course cosine X and this is another cosine X and this is one times that which is we can PA this up as
5:53:45
secant x pair them up we get tangent X very nice right very nice very nice very nice so that's it the derivative of secant x by the definition I'm not going to do the the of tangent by definition because I'll give you guys something better anyway number 97 I will have to um erase all this how are you guys doing are you guys still watching are you guys still doing good let me know all right now number 97 right here no kidding I will show you guys for the first
5:54:42
time on my channel how to differentiate how to find the derivative of the invers sign of X with the definition of derivative here we go this right here is X sorry that's a bad one here we go by definition we take the limit as H approaching zero and now this right here is the inverse sign so we have the inverse right and this inside is x + H and then we minus of course the original function which is the inverse s of X like that you don't put
5:55:23
a plus h and over h right here is the hardest part it's like how in the world can we simplify the top well we just have to do the angle addition identity kind of like backwards I will show you what I mean by that so let me just make a note right here this time though we want to look at the angle difference formula because whenever we are taking the infers trick function it represents an angle right so you see we have the the first angle minus
5:55:58
the second angle that's what we have to look at so in fact I will have to look at sign of a minus B these two angles are being subtracted okay so that's the key well if you look at this run right here let me tell you the identity right here is sin a cosine B and you minus sin B cosine a and usually I use Alpha Beta but you know why so serious ious about like this right here 97 already okay here is the deal this right here it's angle I will put it to be the inverse s of x +
5:56:40
H uh hold on let's let's do this right here let's do this right here okay so we're good with this identity right and now you see this right here sign of a minus B so we are just going to take the inverse sign on both sides take the inverse sign here and take the inverse sign right here so that this and that will cancel out and don't worry about the domain because right here everything is small right here right see X Plus H and um we can yeah just
5:57:20
just just negative pi over two 2 pi that's that's good anyway so this is this is good and uh you will see that right here we can come up with a minus B and this right here is going to give us inverse sign of sin a cosine B okay and then minus sin B cosine a so I think this is believable now what I'm going to do is I'm going to let a equal to inverse sign again this is meant to be an angle right here iners sign of X plus h and of course B will just be
5:58:00
invers sign of X and guess what I'm going to do of course I will just put all the A's right here uh yeah too many arrows though I think I'm not going to draw arrows otherwise it's too confusing right but here is the trouble if I put in for sign of this in here yes it's easy they can just cancel out like this but for the coine it does not cancel out nicely it's okay though this is what we can do of course we all know uh sin square of theta
5:58:38
plus cosine Square Theta is equal to 1 well we can move this to the other side and we can just take the square root so in other word cosine Theta is equal to square root of 1 minus sin Square Theta right okay so you might be wondering should we put on the plus minus um technical we do but let's let's just do the positive version first okay anyway so what I'm going to do is right here what I'm going to do right here is in first sign and this right here is the same as saying
5:59:25
sin a and instead of cosine B we we can look at this as the square root of 1 minus sin square B right just like this and again let's just deal with the Positive version first uh the way you can do it is that you can just say that x equal to uh that X approaching zero plus and you can say that's posi positive I think that's how you can do it where you can just restrict the domain and then the other one should be symmetrical minus minus and you can be able to factor out all that
5:59:57
stuff I think that's will be it so let's just do this part and then the other part you guys do it anyway for the other one minus sin B and this is cosine a we don't like that we write it as square root of 1 minus s square a so this is the fion that I'm going to use now let's do this right here come back here and we take the limit as h approaching zero let's see what we have right here well well I am going to put this right here is the a put it in here so I will just do it one
6:00:43
by one for you guys okay put it one here all right when we do that you see sign and inverse will cancel that means we just have X Plus H left right here first and don't forget we still have the inverse sign right here so I will write it down for you guys we still have the big inverse sign right here in the front and the first thing that we get right here is X Plus H only so I will just put in X Plus H right and then we'll do this put it here this is the B put it here and we have
6:01:17
the square root sign and iners sign will cancel so you end that it's just X right here and to the second power so you multiply this by 1 minus x s like that all right and then continue uh you do the same we do the same thing this time this right here put it here it's just X because sign and in sign cancel so it's just X so that's a minus by the way so we have minus just X and then lastly we put this in here and pretty much the same speech we get square root of 1 minus
6:01:56
parentheses x + h8 H square like that okay and then this is the all over H so this is how you play around with the angle difference formula right so check this out we still have the iners sign or another stuff right are we good hopefully hopefully right now we're all good but the next one I know a lot of you guys will not be so happy about it but but I'm going to do it anyway recall recall the limit as any angle Theta okay it can be H but let me just use any angle Theta sin Theta
6:02:40
over Theta this right here is equal to one okay here is the deal if you would like you can look at this as the limit as Theta approaching zero of sin Theta this being equal to the Limit as Theta approaching zero of just Theta you can look at this as the limit of the top over the limit of the bottom and then multiply limit of uh that on both sides divide the same thing okay yeah lastly what I'm going to do is based on this I will tell you the limit of uh
6:03:23
Theta approaching zero I'm going to take the inverse sign on both both sides and of course they continuous so so we can do that and when we take the infers on both sides that means the limit as data approaching zero of theta this right here is equal to the Limit as Theta approaching zero of inverse s data like this so what am I doing well look at this one and look at the backwards what's saying is that hey if the inside right here
6:03:54
is small even though you're inside of the inverse sign you can forget about it check this out X so H approaching zero if you putting H approaching zero here you're putting zero this is just actually the same as that so inside it's actually approaching zero as well so this is legitim let me put this down this right here is a limit as H approaching zero you can just drop the inverse sign because again we have the
6:04:22
inside is approaching zero so we can just put this down X plus h square root of 1 - x s and then this is minus x square root of 1 minus excuse me because it's being so many LS right and then this is all over H right so again that and now seriously you don't have to do you you should not do Lal rule on this you should do geometry for this and then based on that you can come with all this and the picture is like this okay this is the picture so you get
6:05:04
this right here you can get the triangle and you can get the arc and you can get a uh tangent and you can you know do some series room to figure that out 100 limit I'll promise you I'll do that in my 100 limit video but not now so hopefully this is the jet okay okay this is pretty good because this is pretty algebraic there's no more infers sign with regular sign whatsoever and the good thing is if this is alge breaking now we can just do the usual thing right do the
6:05:45
conjugate so multiply this right here with the Bottom by the conjugate namly we just do X plus h * s < TK of 1 - x 2+ x s < TK of 1 + x + H squ this right here and also put that one on the bottom right here squ < TK of 1 - x s + x s < TK of 1 + parenthesis x + H squ like that man this is life on the top this actually not that bad to multiply the out because it's in the form of a minus B A plus b you guys should know that right here right by here we go so we
6:06:37
will just have this being squared namely x + H being squar and this thing Square which is just 1 - x squ okay and then minus this and that so you pretty much just have X being squar and this right here being squared by itself which is 1 + parentheses x + h² like that and uh multiply out the bottom okay so here we have no don't multiply out the bottom and then just I mean rewrite the bottom x + H and square < TK of 1 - x² + x square < TK of 1 + x + H squar yeah
6:07:32
man why is this so let's see is there any fast way to cancel this guy out I feel this should be a fast way to cancel the top let's see there I think is all right so just a top okay just the top just the top and just the top because I don't want to write this guy down again so if you take this guy and then distribute it okay so let's see what we get we should get just the top equal sign just algebra so X Plus H squar yeah and then minus x² * that guy so
6:08:18
x s * x + H Square yeah next we take this guy distribute it distribute so we have Negative X squ and then minus X squ ah yes yes I see this guy now okay okay okay okay we're good we're good we're good x square * that guy so x sare * x + H square like that okay you kidding me this is minus wait they don't cancel all [Music] nicely let's see let see let see did I make a mistake somewhere hopefully
6:09:07
not I am afraid so I didn't make a mistake okay 1 - x s this should have been one minus oneus okay one minus because remember it's oneus cosine square and then you take cosine to be the inverse all that stuff right sorry oneus s Square so that should have be a minus so this right here should have been a minus this right here should have been a minus this right here should have been a minus this right here should have been a minus
6:09:58
so you see this times this is okay negative times this guy it becomes a Plus thank all my subscribers all right so okay hopefully you guys okay okay if you want to watch how to do this of course you should be able to do all that anyway this and that cancel each other nicely and have been uh what we did this that's let me not do anything F me just multiply this out this is x² + 2xh + h² and this is - x s so this and that cancel again and you see I can Factor the H so we get 2x
6:10:41
+ H like that this is what happen on the top all right so I will just do this one legally okay this right here is the limit as H approaching zero this is on the top all right so we have H time 2x + H okay and uh we have the H all the way in the front okay here we go X Plus H and then this is the square root of 1 - x s and plus x one square root of 1 me this minus and then x + H squ like this oh my my God cancel out this H oh cool and now of course we just have to plug in zero into
6:11:33
all the H so putting zero in here we get 2x + 0 over okay don't panic this right here is just x + 0 this right here is square root of 1 - x 2 and then we add x * < TK of 1us parentheses x + 0 squ yeah man I also want I just want to put a 2X plus Zer finally we see on the top is just 2x on the bottom check this out this right here is x times that x * the same thing so you have two of them namely you have 2x times what s root of 1 - x s and what happen with the 2x bye-bye bye-bye
6:12:22
we get 1 over square otk of 1 - x s all oh my goodness how long did this take me oh my God man so as I told you this is why I said oh man I don't know if I was ready for this or not but this is it this is it okay this is hopefully from here to here you guys are feeling okay you guys think this is really legit and all that right so good 10 o' already man all right number [Laughter] 97 in first tent effect with the definition all right so without wasting any
6:13:23
time number 98 D DX inverse tangent of X right so here we go take the limit as H approaching zero and again this right here is H so this is this is X this right here is going to be the inverse tangent of x + H and then minus inverse tangent of X and then this is all over H so what do we do we do the same thing as what we did earlier therefore right here we will just do the tangent situation and again want the tangent subtract the angle subtraction
6:14:01
formula so you look at tangent of a minus B this right here is equal to let me just tell you guys what the answer is okay this right here is equal to tangent of a plus okay this right here should be a plus sorry tangent of a minus minus right minus tangent of B over one here's plus tangent a * tangent B okay and you know the do what we do is we'll just take the inverse tangent on both sides okay T the inverse tent on both s so they will be gone and then of course for this a we'll
6:14:49
just make it to equal to inverse tangent of x + H and then for the B we just make equal to inverse tangent of X and it's pretty good because we can just draw in everything and then we will be happy with it so you will see that right here this the top okay we still write down the limit as H approaching zero okay put this right here right here don't forget we still have the inverse tangent but you know what we are going to do yes that's later though
6:15:30
inverse tangent tangent a if a is this of course tangent and inverse tangent they cancel namely we just get x + H so on the top here we get X Plus H I'll put it down in blue X Plus H and then we have the minus and of course we put this right here for that so we have minus and and this in here tangent invers tangent will cancel we'll just have minus X over same thing so it's just going to be one plus uh this right here will give us just X Plus H and lastly this part will just
6:16:16
give us X so that's the beauty and I think you guys can see what is going on already right and this right here is all over H this right here is all over H yes I will tell you the limit as uh angle Theta goes to zero of tangent Theta over Theta this right here will also give you one and you can do the same trick like what I show you earlier again you can do this geometrically what I want to tell you is that we can just drop the inverse tangent
6:16:58
legitimately right it's a legitimate work so here we have the limit as H approaching zero drop the inverse tangent but perhaps let me just write the one/ H like this times the blue part right here so no more iners tangent because of this right here okay so here we have x + H oh can just cancel them so just have H right here so just the blue part here so just the H over 1 + parentheses x + H time x yes yes H H cancel out and then you put the zero in here so ladies
6:17:45
and gentlemen in the end we get 1 over 1+ this right here if you put zero here you get get zero you just get x * the other X you get x²
6:17:58
much better right all right couple more questions and we'll be done of course you guys can see I haven't eaten anything yet but yeah all right so the last two we are going to prove the product rule and also the the quo rule so here we go by definition of course 99 D DX just two random gener like not random generating function General functions like two general functions I don't know what they are anyway enough talking here is the
6:18:46
de limit as H approaching zero this is X they are the input so first part we have f of x + H time G of x + H right and then minus f of x time G of X all over h Huh how in the world can we prove this guy well this is one of the example that you have to add something and a subtra something to make things work let me show you of course we have to know the derivative of f and g that they do have to exist remember what does it mean to have a derivative the
6:19:31
Der the derivative the definition of derivative frime of x this right here is the limit as H approaching zero f of x + h f of x + H minus F ofx all over h like that yeah that's the definition of frime of X and that's the one we have been doing um already well H I somehow see that we have the f of x plus h let's see if we can do the following to make this workout nicely or not this right here is the limit as X approaching zero we have the H
6:20:27
right here this right here it's kind of similar to the idea when we're doing the derivative sign you want to kind of factor out the the ones the the function part that doesn't have the H this right here has the f of x h and also want to have the G to be involved this is what we are going to do perhaps I will have f of x plus H * G of x + H okay this right now is looks pretty symmetric but what I would like to do is perhaps let me minus f of x + H and you see when
6:21:09
you do that you can be able to factor this out well in the meantime that though I would like to just multiply by G of X without the H and the idea is kind of like you can just pick this with that this with that and this also a geometry proof that you can check that out you can check out Dr P video for that but it's been 6 hours and 20 something minutes so let me just finish this so I can go home thank you okay right here you cannot just subtract something you also
6:21:38
have to add it back so you add it with f of X+ H time G of X so this is still equivalent than the original and then minus f of x g of X and you can see that it will be a pretty well balanced thing to do okay I'm just going to look at the first part and keep in mind these guys they all have the H in the denominator so look at this guy with h and is there anything that we can do well well let's see both of them have the f of x plus h I'm going to factor it out but I cannot
6:22:18
take it out to the uh limit so I could just you know stick with it the limit as H approach the limit as H approaching zero let me just Factor this guy out which is f of x + H and that's it times the limit perhaps I'll put this down in red I still care about the color the limit as H approaching zero I'll do this minus that over that and you see that will be the G Prime if you just change off the F to G from here that's very nice so let me put down G of x + H minus G of X
6:23:01
or over H so if you just you know this part just Factor this out and then write this down that's the first part that's very good now we add to do the same limit but for this one we can factor out the G of X and this one right here you don't have to do this because it's just G of X you can put it out side of the limit without any worry and then the left over is exactly what we have right here so we multiply by the limit as H approaching zero and this minus that so f of x + H minus
6:23:41
f of x all over H now this right here F has to be continuous otherwise you cannot have limit right so you can just put put the zero right here nicely and just f of x plus Zer just f of x actually yeah so it's it's all good this right here is just F ofx namely the original function times this right here is the definition of g Prime so it's just G Prime of X this right here plus G Prime of no this right here it's just G Prime this right here just G of X this right here though is
6:24:21
frime of X so we multiply by frime of X I guess guess what we are done ladies and gentlemen the product prod okay the prodct man I've been saying product rule for the whole video feel so bad now I am wondering if I should do the next one will not which is the quo rule what should you guys do it okay I what do this I will do this okay number 100 D DX f of x over G of X you guys do it this
6:25:08
right here will be the bottom one square and then the bottom one times the Der to the top one which is frime of X and then minus the top one which is f ofx times the derivative of the bottom one this right here is it well uh the the hard part is you have to figure out what to add what to subtract and well just just stud up you you you guys can do it get a common denom just multiply the top and bottom by the common denominator you end up with the complex fractions I hope you
6:25:49
guys are not so mad because I am not doing that one I don't know I feel so bad if I don't do it all I will do it I do it I do it I do I do I do man okay ignore that number 100 number 100 right here we are going to be proving the quo rule D DX and f of x g of X and dividing here we go go ahead write down the limit as H approaching zero and on the top here we have F of you see the input is X and X here so
6:26:42
that means we have f of x + H over G of x + H and then minus the original namely f of x over G of X or over H like that well well let's just multiply let's just fix the what's that called the complex fraction situation first so multiply the top and bottom by the denominator which is G of X plus h g of X and G of X plus h g of X so you gu can see this is rather similar to the secant right so so that's why I've been putting down by organize all the work sheet right here so this time
6:27:31
I'm going to just put that as a limit by itself for the denominator right here because you guys have seen the secant situation right hopefully so one over this and that okay just the blue portion G of X like this one over all right now for the part for the left over this is the limit as H approaching zero okay when you multiply this and that this and that will cancel so you have this following f
6:28:10
of x + H time G of X yeah this times that and then minus f of x and uh times this which is G of x + H like this let's see if it's correct is correct and all over H right so that's the one that we have and now perhaps let's take care of this part first right here you can see that if we put zero into H what do we get this will give us zero here so we end out as G of x * G of X namely we get 1 over g s Nam the the Der the bottom one and you square right so that's
6:29:08
that part and now for this one of course you know what's coming you are going to take the limit as H approaching zero and you going to find something to add and find something to subtract well the same thing to subtract Okay so H right here hopefully it should be enough space this right here f of x + H and G of X right and this right here is minus f of x g of x + H yeah now uh I want to have the derivative come up to help us out right so remember you should be able
6:29:53
to factor things out and all that hm what should we do here we are going to minus um you might want to try to factor out the G of X first so I want to G of X in the meantime if you want to have the derivative form you want to have f ofx plus h minus f of x times G of X so this way you can Factor the G of X don't just add don't forget to I mean don't just subtract don't forget to uh add the back so F of of x g of X this is what you have to do now it's just how we write it down
6:30:32
nicely this right here is 1 over G of X and we are going to square that so perhaps it should still keep this in blue or not 1 over G of x square okay so we'll do this not your head like but I write it down I will factor out the G of X okay so this is the parenthesis we factor out the G of X let just St red man this times that write it down as G of X take the limit of the remaining which is the
6:31:09
limit as H approaching zero and we have f of x + H minus f of x all over H right next part we add
6:31:28
why is this adding why is this adding uh oh f of x oh don't worry I almost have a heart attack okay okay okay we're good we're good we're good good okay right here uh F ofx and F ofx we can factor out so we still add okay for now still add this right here is g no this right here is f of x f of x right here okay Factor the out and then we have the limit as H approaching zero now this is G of of x minus G
6:32:28
of x + H this is something kind weird yes okay this right here is not what we preferred it's okay just factor out negative so we can get G of X Plus H minus G of X this negative man scared me all right all all let's write down the top right here right so we have G of x times this right here is the derivative right so derivative of the top and then this right here right this a minus right plus time minus so it's a minus and we have the function f ofx and this part right
6:33:10
here is G Prime of X all in all divided by the bottom function square and we're done this is 100 derivative but you guys know I have another one for you guys this right here it's the adult power rule I will try to fit in here right so I will try to fit in here ladies and gentlemen 101 this right here is the adult power rule meaning that you put the three right here titration okay titration okay don't worry what does this mean I will tell you this means okay if you have the
6:33:59
three right here it means x * x * X but when you have this three right here this means we differentiate x to the X to the X power like that right well again we have the base is X we don't like that so we are going to look at the derivative well this right here is just e to the natural log of x raise to the X to the X power but we can put that in the front so it's just X to X right here like this and now I will write this down for you we can differentiate this guy
6:34:32
already um perhaps I will just I just do here let's I think I should be okay why the already so big all right again this is e to the natural log of x and then this is X tox right okay here we go e to however we just get the same thing right e to the that whatever times Well product rule in action first one we have x to the X that's the first function and we multiply by the to the second which is 1 /x and then we add the second function which is natural log of x but
6:35:13
do you guys remember what's the derivative of x to the X maybe you have to go back to watch the video huh I will tell you is X to the X but this kind weird like going down like this times uh let me I see it's 1 plus natural log of x like this okay so no kidding now this right here is what well if you look back it's just X to X to X so this is X to x to the x x to X to X and notice this guy has X tox and this guy also has X to thex so I'm going to factor that out right here
6:35:49
as well okay right here I have 1/x right here I have natural log of x perhaps we will distribute this in so we get plus natural log of x right and then plus natural log of x * natural log X ladies and gentlemen Lal log of x square done oh my God take a look take a look take a look oh my God I did 101 derivative this is 6 hours and 30 minutes the longest video I have done so far this is man w w right thank you guys so much for watching and again download the file
6:36:38
and I have the other file for you guys as well and of course I'm going to put down my M met for the uh uh 100 derivatives so funny this right here I was I pr did this uh 4 hours and 22 minutes what so yeah that was when I was much younger and this right here also PR did this right here 4 hour I mean 6 hour and 37 minutes oh my God this is just amazing all right so hopefully you guys all find this video to be
6:37:25
helpful and leave a comment down below and let me know if you guys have any questions and yeah uh give me a like thank you guys so much if you guys enjoy this video just give me a like thank you and then again this video is dedicated to my uh upper bound students especially summer 2009 at UC berkly campus right and Mr monik LBA thank you so much for giving me the chance to become a teacher and okay I love my girlfriend J and of course man this is so wonderful whenever I finish
6:38:02
this yeah okay as always that's it
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