# How do you differentiate f(x)=e^(-2x^2+x+8)  using the chain rule?

Mar 17, 2016

$f ' \left(x\right) = \left(1 - 4 x\right) {e}^{- 2 {x}^{2} + x + 8}$

#### Explanation:

The chain rule states that derivative of a function of a function, say

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dg}} \cdot \frac{\mathrm{dg}}{\mathrm{dx}}$

So if $f \left(x\right) = {e}^{- 2 {x}^{2} + x + 8}$, (here $f \left(x\right) = {e}^{g} \left(x\right)$,

where $g \left(x\right) = \left(- 2 {x}^{2} + x + 8\right)$. Hence

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{\mathrm{de}}^{- 2 {x}^{2} + x + 8}}{d \left(- 2 {x}^{2} + x + 8\right)} \cdot \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} + x + 8\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = {e}^{- 2 {x}^{2} + x + 8} \cdot \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} + x + 8\right)$

= ${e}^{- 2 {x}^{2} + x + 8} \cdot \left(- 4 x + 1\right)$

= $\left(1 - 4 x\right) {e}^{- 2 {x}^{2} + x + 8}$