How do you differentiate p(y) = y^2sin(y) using the product rule?

Dec 1, 2015

$p ' \left(y\right) = 2 y \sin \left(y\right) + {y}^{2} \cos \left(y\right)$

Explanation:

According to the product rule:

$p ' \left(y\right) = \sin \left(y\right) \frac{d}{\mathrm{dy}} \left[{y}^{2}\right] + {y}^{2} \frac{d}{\mathrm{dy}} \left[\sin \left(y\right)\right]$

Find each derivative separately.

$\frac{d}{\mathrm{dy}} \left[{y}^{2}\right] = 2 y$

$\frac{d}{\mathrm{dy}} \left[\sin \left(y\right)\right] = \cos \left(y\right)$

Plug them back in:

$p ' \left(y\right) = 2 y \sin \left(y\right) + {y}^{2} \cos \left(y\right)$

If you want to factor:

$p ' \left(y\right) = y \left(2 \sin \left(y\right) + y \cos \left(y\right)\right)$