Chain rule with the power rule

So we've got the function ( f(x) = (2x^3 + 5x^2 - 7)^{88} ) and we want to find the derivative of our function ( f ) with respect to ( x ). Now, the key here is to realize that this function can be viewed as a composition of two functions. How do we do that? Well, let me diagram it out.

So let's say we want to start with—I'll do it down here so I have some space. We're going to start with an ( x ), and what's the first thing that we would do if you were just trying to evaluate it given some ( x )? Well, first, you would take ( 2 \cdot x^3 + 5 \cdot x^2 - 7 ).

So what if we imagined a function here that just did that first part, that just evaluated ( 2x^3 + 5x^2 - 7 ) for your ( x )? Let's call that the function ( U ). Whatever you input into that function ( U ), you're going to get ( 2 \cdot ) (that input) to the 3rd power ( + 5 \cdot ) (that input) to the 2nd power ( - 7 ).

And so when you do that, when you input with an ( x ), what do you output? What do you output here? Well, you're going to output ( U(x) ), which is equal to ( 2x^3 + 5x^2 - 7 ).

Now, what's the next thing you're doing? You're not done evaluating ( f(x) ). You would then take that value and then input it into another function. You would then take the 88th power of that value. So then we will take that and input it into another function—let's call that function ( V ).

And that function, whatever input you give it, and I'm using these squares just to say whatever input goes into that function, you're going to take it to the 88th power. And so in this case, what do you end up with? Well, you end up with ( V(U(x)) ) or you could view this as ( V(2x^3 + 5x^2 - 7) ) or you could view this as ( (2x^3 + 5x^2 - 7)^{88} ).

And that's what ( f(x) ) is. As we just saw, ( f(x) ) can be viewed as the composition of ( V ) and ( U ). This is ( f(x) ).

So if we write ( f(x) ) being equal to ( V(U(x)) ), then we see very clearly that the chain rule is very useful here. The chain rule tells us that ( f'(x) ) is going to be the derivative of ( V ) with respect to ( U ), so it's going to be ( V'(U(x)) ) times the derivative of ( U ) with respect to ( x ), so ( U'(x) ).

So we know a few things already, so let me just write things down very clearly. We know that ( U(x) = 2x^3 + 5x^2 - 7 ). What is ( U'(x) )? Well, here we're just going to use some derivative properties and the power rule.

( 3 \cdot 2 = 6 ), ( x^{3-1} = x^2 ), so ( U'(x) = 6x^2 ). ( 2 \cdot 5 = 10 ), take one off that exponent; it's going to be ( 10x ). And the derivative of a constant is just zero, so we can just ignore that. So that's ( U'(x) ).

Now, we know that ( V )—if we input an ( x ) into ( V )—so ( V(x) = x^{88} ). ( V'(x) )? Well, we just use the power rule again; that's ( 88x^{87} ).

So ( V'(U(x)) ) — if you were to input ( U(x) ) into ( V' ) — well, it's going to be equal to ( 88 \cdot (U(x))^{87} ). Whatever you input into ( V' ), you're going to take it to the 87th power and multiply it by 88.

So ( U(x) ) and that's the same thing; this is equal to ( 88 \times ) (this entire expression) ( U(x) = 2x^3 + 5x^2 - 7 ).

So there you have it: ( f'(x) = V'(U(x)) ) is all of this business, so it's equal to ( 88 \cdot (2x^3 + 5x^2 - 7)^{87} \times U'(x) ).

We figured out ( U'(x) ), so times ( (6x^2 + 10x) ). Now as you get more practice with the chain rule, you'll recognize this faster and actually, you could do it in your head. You’ll say, "Okay, I’m going to take the derivative of the outside function"—the blue function you could say—with respect to what I have on the inside.

So if I was taking the derivative of ( x^{88} ), it would be ( 88x^{87} ) if that’s with respect to ( x ). But if I'm taking the derivative of this with respect to the inside, well, where I had the ( x )s before, I would just have this ( U(x) ), so it's going to be ( 88 \times (U(x))^{87} ) and I multiply that times the derivative of the inside which is ( (6x^2 + 10x) ).