# Question #56c3f

Nov 7, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \cos \left({x}^{2} + 1\right) {e}^{- 2 x} \left(2 x \sin \left({x}^{2} + 1\right) + \cos \left({x}^{2} + 1\right)\right)$

#### Explanation:

$y = {e}^{- 2 x} {\cos}^{2} \left({x}^{2} + 1\right) = u v$

If $y = f \left(x\right) g \left(x\right) \implies \frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$

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

$g \left(x\right) = {\cos}^{2} \left({x}^{2} + 1\right) = {\left(\cos \left({x}^{2} + 1\right)\right)}^{2}$
$g ' \left(x\right) = 2 \cos \left({x}^{2} + 1\right) \cdot \frac{d}{\mathrm{dx}} \left[\cos \left({x}^{2} + 1\right)\right]$
$g ' \left(x\right) = 2 \cos \left({x}^{2} + 1\right) \cdot - 2 x \sin \left({x}^{2} + 1\right)$
$g ' \left(x\right) = - 4 x \sin \left({x}^{2} + 1\right) \cos \left({x}^{2} + 1\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 4 x \sin \left({x}^{2} + 1\right) \cos \left({x}^{2} + 1\right) {e}^{- 2 x} - 2 {e}^{- 2 x} {\cos}^{2} \left({x}^{2} + 1\right) = - 2 \cos \left({x}^{2} + 1\right) {e}^{- 2 x} \left(2 x \sin \left({x}^{2} + 1\right) + \cos \left({x}^{2} + 1\right)\right)$