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

May 4, 2018

$\frac{d}{\mathrm{dx}} \left({e}^{{x}^{2} + x + 8}\right) = \left(2 x + 1\right) {e}^{{x}^{2} + x + 8}$

#### Explanation:

Based on the chain rule, let $y \left(x\right) = {x}^{2} + x + 8$, then:

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

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y} \left(2 x + 1\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x + 1\right) {e}^{{x}^{2} + x + 8}$