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

Aug 29, 2016

$f ' \left(x\right) = \left(2 x - 3\right) {e}^{{x}^{2} - 3 x - 4}$

#### Explanation:

Differentiate using the $\textcolor{b l u e}{\text{chain rule}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots . \left(A\right)$

let $u = {x}^{2} - 3 x - 4 \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 2 x - 3$

then $f \left(x\right) = y = {e}^{u} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u}$

Substitute these values into (A) and change u back into terms of x.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{u} \left(2 x - 3\right) = \left(2 x - 3\right) {e}^{{x}^{2} - 3 x - 4}$