# How do you differentiate p(y) = e^(y^2)+sin^2(y)cos(y)-ye^(y) using the product rule?

Mar 12, 2017

$\frac{\mathrm{dp}}{\mathrm{dy}} = 2 y {e}^{{y}^{2}} + 2 \sin y {\cos}^{2} y - {\sin}^{3} y - {e}^{y} - y {e}^{y}$
$p \left(y\right) = {e}^{{y}^{2}} + {\sin}^{2} y \cos y - y {e}^{y}$
Hence, $\frac{\mathrm{dp}}{\mathrm{dy}} = {e}^{{y}^{2}} \times 2 y + \left(2 \sin y \cos y \times \cos y + {\sin}^{2} y \times \left(- \sin y\right)\right) - \left(1 \times {e}^{y} + y {e}^{y}\right)$
= $2 y {e}^{{y}^{2}} + 2 \sin y {\cos}^{2} y - {\sin}^{3} y - {e}^{y} - y {e}^{y}$