Applying the chain rule and product rule | Advanced derivatives | AP Calculus AB | Khan Academy

What we're going to do in this video is try to find the derivative with respect to X of (x^2 \sin(X)) all of that to the third power.

And what's going to be interesting is that there are multiple ways to tackle it. I encourage you to pause the video and see if you can work through it on your own. So, there's actually multiple techniques. One path is to do the chain rule first, so I'll just say CR for chain rule first.

So, I have I'm taking the derivative with respect to X of something to the third power. If I take the derivative, it would be the derivative with respect to that something, so that would be (3 \times) that something squared times the derivative with respect to X of that something, where the something in this case is (x^2 \sin(X)).

This is just an application of the chain rule. Now, the second part, what would that be? The second part here, just in another color, in orange, well here I would apply the product rule. I have the product of two expressions, so I would take the derivative of... let me write this down.

So this is going to be the product rule. I would take the derivative of the first expression. The derivative of (x^2) is (2x). Write a little bit to the right, it's going to be (2x \times) the second expression (\sin(X) +) the first expression (x^2 \times) the derivative of the second one (\cos(X)).

That's just the product rule as applied to this part right over here, and all of that, of course, is being multiplied by this up front, which actually, let me just rewrite that.

So, all of this I could rewrite as... let's see, this would be (3 \times) if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. So (x^2) squared is (x^4) and (\sin^2(X)) is (\sin^2(X)), and then all of that is being multiplied by that.

And if we want, we can algebraically simplify. We can distribute everything out, in which case, what would we get? Well, let's see. (3 \times 2) is (6), (x^4 \times x) is (x^5), (\sin^2(X) \times \sin(X)) is (\sin^3(X)), and then let’s see, (3 \times x^4 \times x^2) is (x^6), and then I have (\sin^2(X) \cos(X)).

So there you have it. That's one strategy: chain rule first and then product rule. What would be another strategy? Pause the video and try to think of it. Well, we could just algebraically use our exponent properties first, in which case this is going to be equal to the derivative with respect to X of... if I'm taking (x^2 \cdot \sin(X)^3) instead, I could say (x^2) raised to the third power, which is going to be (x^6) and (\sin(X)) raised to the third power.

I'm using the same exponent property that we used right over here to simplify this. If I'm taking the product of things to some exponent, well, that's the same thing as each of them raised to the exponent and then the product of the two.

Now, how would we tackle this? Well, here I would do the product rule first, so let's do that. We're going to take the derivative of the first expression, so the derivative of (x^6) is (6x^5) times the second expression (\sin^3(X)) plus the first (x^6) times the derivative of the second.

And I'm just going to write that as (D(X) = \sin^3(X)). Now, to evaluate this right over here, it does definitely make sense to use the chain rule.

So, what is this going to be? Well, I have the derivative of something to the third power, so that's going to be (3 \times) that something squared times the derivative of that something. So in this case, the something is (\sin(X)) and the derivative of (\sin(X)) is (\cos(X)).

And then I have all of this business over here. I have (6x^5 \sin^3(X) + x^6), and if I were to just simplify this a little bit, in fact, you see it very clearly these two things are equivalent. This term is exactly equivalent to this term the way it's written.

And then this is exactly, if you multiply (3 \times x^6 \sin^2(X) \cos(X)), so the nice thing about math is if we're doing things that make logical sense, we should get to the same endpoint.

But the point here is that there are multiple strategies you could use: the chain rule first and then the product rule, or you could use the product rule first and then the chain rule.

In this case, you could debate which one is faster. It looks like the one on the right might be a little bit faster, but sometimes these two are pretty close.

But sometimes it'll be more clear than not which one is preferable. You really want to minimize the amount of hairiness, the number of steps, and the chances for careless mistakes you might have.