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

##### 1 Answer
Feb 15, 2016

$f ' \left(x\right) = - 4 x {e}^{- 2 {x}^{2} - 4}$

#### 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

$\frac{d}{\mathrm{dx}} \left[g \left(h \left(x\right)\right)\right] = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$

Here, we have $g \left(x\right) = {e}^{x}$ and $h \left(x\right) = {x}^{2} - 3 {x}^{2} - 4$ which can be rewritten as $h \left(x\right) = - 2 {x}^{2} - 4$.

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

${\overbrace{{e}^{- 2 {x}^{2} - 4}}}^{g ' \left(h \left(x\right)\right)}$

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

Thus,

$f ' \left(x\right) = {\overbrace{{e}^{- 2 {x}^{2} - 4}}}^{g ' \left(h \left(x\right)\right)} \cdot {\overbrace{\left(- 4 x\right)}}^{h ' \left(x\right)}$