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  • - [Voiceover] What we want to do is find the

  • derivative of this G of X

  • and at first it can look intimidating.

  • We have a sine of X here.

  • We have a cosine of X.

  • We have this crazy expression here

  • with a pi over cube root of X

  • we're squaring the whole thing

  • and at first it might seem intimidating.

  • But as we'll see in this video, we can

  • actually do this with the tools already in our tool kit.

  • Using our existing derivative properties

  • using what we know about the power rule

  • which tells us the derivative with respect to X.

  • Of X to the N

  • is equal to N

  • times X to the N minus one,

  • we've see that multiple times.

  • We also need to use the fact

  • that the derivative of cosine of X

  • is equal to negative the sine of X.

  • And the other way around

  • the derivative with respect to X

  • of sine of X

  • is equal to positive

  • cosine of X.

  • So using just that we can actually evaluate this.

  • Or evaluate G prime of X.

  • So, pause the video and see if you can do it.

  • So probably the most intimidating part of this

  • because we know the derivative's a sine

  • of X and cosine of X is this expression here.

  • And we can just rewrite this

  • or simplify it a little bit so it takes a form

  • that you might be a little bit more familiar with.

  • So, so let me just do this

  • on the side here.

  • So, pi

  • pi over

  • the cube root of X

  • squared.

  • Well that's the same thing.

  • This is equal to pi squared

  • over the cube root of X squared.

  • This is just exponent properties

  • that we're dealing with.

  • And so this is the same thing.

  • We're gonna take X to the one third power

  • and then raise that to the second power.

  • So this is equal to pi squared

  • over.

  • Let me write it this way,

  • I'm not gonna skip any steps

  • because this is a good review of exponent property.

  • X to the one third squared.

  • Which is the same thing as pi squared

  • over X to the two thirds power.

  • Which is the same thing as pi squared

  • times X to the negative two thirds power.

  • So when you write it like this

  • it starts to get into a formula,

  • you're like, oh, I can see how the power rule

  • could apply there.

  • So this thing is just pi squared

  • times X to the negative two thirds power.

  • So actually let me delete this.

  • So,

  • this thing

  • can be rewritten.

  • This thing can be rewritten

  • as pi squared

  • times X to the negative

  • to the negative two thirds power.

  • So now let's take the derivative

  • of each of these pieces of this expression.

  • So, we're gonna take

  • we want to evaluate what the G prime of X is.

  • So G prime of X

  • is going to be equal to.

  • You could view it as a derivative with respect

  • to X of seven sine of X.

  • So

  • we can take do the derivative operator

  • on both sides here just to make it clear

  • what we're doing.

  • So we're gonna apply it there.

  • We're gonna apply it there.

  • And we're going to apply it

  • there.

  • So this derivative

  • this is the same thing as

  • this is going to be seven times the derivative

  • of sine of X.

  • So this is just gonna be seven times

  • cosine of X.

  • This one, over here,

  • this is gonna be three, or we're subtracting,

  • so it's gonna be this subtract

  • this minus.

  • We can bring the constant out

  • that we're multiplying the expression by.

  • And the derivative of cosine of X

  • so it's minus three times

  • the derivative of cosine of X

  • is negative sine of X.

  • Negative sine of X.

  • And then finally

  • here in the yellow we just apply the power rule.

  • So, we have the negative two thirds,

  • actually, let's not forget this minus sign

  • I'm gonna write it out here.

  • So you have the negative two thirds.

  • You multiply the exponent times the coefficient.

  • It might look confusing, pi squared,

  • but that's just a number.

  • So it's gonna be

  • negative

  • and then you have negative two thirds

  • times pi squared.

  • Times pi squared.

  • Times X to the negative two thirds

  • minus one power.

  • Negative two thirds

  • minus one power.

  • So what is this going to be?

  • So we get G prime of X

  • is equal to

  • is equal to

  • seven cosine of X.

  • And let's see, we have a negative three

  • times a negative sine of X.

  • So that's a positive three sine of X.

  • And then

  • we have, we're subtracting

  • but then this is going to be a negative,

  • so that's going to be a positive.

  • So we can say

  • plus two pi squared over three.

  • Two pi squared over three.

  • That's that part there.

  • Times X

  • to the.

  • So negative two thirds

  • minus one,

  • we could say negative one and two thirds,

  • or we could say negative five thirds by power.

  • Negative five thirds power.

  • And there you have it.

  • We were able to tackle this thing that

  • looked a little bit hairy

  • but all we had to use was the power rule

  • and what we knew to be the derivatives

  • of sine and cosine.

- [Voiceover] What we want to do is find the

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Worked example: Derivatives of sin(x) and cos(x) | Derivative rules | AP Calculus AB | Khan Academy

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    yukang920108 に公開 2022 年 07 月 12 日
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