# How do you differentiate f(x)=e^(4x)*sin(5-x^2) using the product rule?

Dec 4, 2015

$f ' \left(x\right) = 2 {e}^{4 x} \left(2 \sin \left(5 - {x}^{2}\right) - x \cos \left(5 - {x}^{2}\right)\right)$

#### Explanation:

$f ' \left(x\right) = \sin \left(5 - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right] + {e}^{4 x} \frac{d}{\mathrm{dx}} \left[\sin \left(5 - {x}^{2}\right)\right]$

Find each individual derivative, both of which use the chain rule:

$\frac{d}{\mathrm{dx}} \left[{e}^{4 x}\right] = {e}^{4 x} \frac{d}{\mathrm{dx}} \left[4 x\right] = 4 {e}^{4 x}$

$\frac{d}{\mathrm{dx}} \left[\sin \left(5 - {x}^{2}\right)\right] = \cos \left(5 - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left[5 - {x}^{2}\right] = - 2 x \cos \left(5 - {x}^{2}\right)$

Plug back in:

$f ' \left(x\right) = 4 {e}^{4 x} \sin \left(5 - {x}^{2}\right) - 2 x {e}^{4 x} \cos \left(5 - {x}^{2}\right)$

$f ' \left(x\right) = 2 {e}^{4 x} \left(2 \sin \left(5 - {x}^{2}\right) - x \cos \left(5 - {x}^{2}\right)\right)$