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


3m read
·Nov 11, 2024

What we want to do is find the derivative of this G of X. At first, it could look intimidating. We have a s of X here, we have a cosine of X, we have this crazy expression here, we have 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 toolkit. Using our existing derivative properties, using what we know about the power rule, which tells us that the derivative with respect to X of x to the N is equal to n * x to the n minus one, we've seen that multiple times.

We also need to use the fact that the derivative of cosine of X is equal to negative 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, 'cause we know the derivative of sine of X and cosine of X is this expression here. 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 actually, let me just do this on the side here.

So, Pi over the cube root of X squared, well that's the same thing. This is equal to pi times K^2 over the cube root of x^2. And this is just exponent properties that we're dealing with. So this is the same thing; we're going to take x to the 1/3 power and then raise that to the second power. So this is equal to pi times K^2 over x to the 2/3 power, which is the same thing as pi^2 over x to the 2/3 power.

So when you write it like this, it starts to get into a form you're like, "Oh, I could see how the power rule could apply there." So this thing is just pi^2 times x to the -2/3 power. So actually, let me delete this.

So this thing can be rewritten as pi^2 times x to the -2/3 power. So now let's take the derivative of each of these pieces of this expression. So we're going to take where we want to evaluate what G Prime of X is. So G Prime of X is going to be equal to, you could view it as the derivative with respect to X of seven sine of X.

So we could take, let's do the derivative operator on both sides here just to make it clear what we're doing. So we're going to apply it there, we're going to 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 going to be 7 times cosine of x.

This one over here, this is going to be three, or we're subtracting, so it's going to be this subtra- this minus. We can bring the constant out that we're multiplying the expression by. The derivative of cosine of x is negative sine of X.

And then finally here in the yellow, we just apply the power rule. So we have the -2/3. Actually, let's not forget this minus sign. I'm going to write it out here. And so you have the -2/3; you multiply the exponent times the coefficient. It might look confusing, pi squared, but that's just a number, so it's going to be negative, and then you have -23 times pi^2 times x to the -2/3 minus one power.

So what is this going to be? So we get G Prime of X is equal to, is equal to 7 cosine of x, and let's see, we have a -3 times a negative sine of X, so that's a positive 3 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 could say plus 2 pi^2 over 3, that's that part there, times x to the -2/3 minus 1. We could say -1 and 2/3, or we could say -5/3 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.

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