Question

How do you differentiate `f(x)=e^(x^2-3x^2-4)` using the chain rule?

Answer 1

`f'(x)=-4xe^(-2x^2-4)`

Explanation:

The chain rule states that when differentiating a function that contains another function inside of it, the following happens: (1) differentiate the outside function and leave the inside function intact. (2) Multiply this by the derivative of the outside function.

The chain rule can be written as

`d/dx[g(h(x))]=g'(h(x))*h'(x)`

Here, we have `g(x)=e^x` and `h(x)=x^2-3x^2-4` which can be rewritten as `h(x)=-2x^2-4`.

Since the derivative of the outside function `g(x)=e^x` is still `g'(x)=e^x`, when we differentiate the outside function and leave the inner portion intact, we will still be left with

`overbrace(e^(-2x^2-4))^(g'(h(x)))`

Which is (the simplified version of) what we began with. Now, we just need to multiply this by the derivative of the inside function, `h(x)=-2x^2-4`. The power rule tells us that `h'(x)=-4x`.

Thus,

`f'(x)=overbrace(e^(-2x^2-4))^(g'(h(x)))*overbrace((-4x))^(h'(x))`