# How do you differentiate p(y) = (y^-1 + y^-2)(5y^-3 + 9y^-4) using the product rule?

Jul 1, 2017

$\frac{\mathrm{dp} \left(y\right)}{\mathrm{dy}} = - 20 {y}^{- 5} - 70 {y}^{- 6} - 54 {y}^{- 7}$

#### Explanation:

According to product rule derivative of $p \left(y\right) = f \left(y\right) g \left(y\right)$ is

$\frac{\mathrm{dp} \left(y\right)}{\mathrm{dy}} = \frac{\mathrm{df}}{\mathrm{dy}} \times g \left(y\right) + f \left(y\right) \times \frac{\mathrm{dg}}{\mathrm{dy}}$

Here we have $f \left(y\right) = {y}^{- 1} + {y}^{- 2}$ and $\frac{\mathrm{df}}{\mathrm{dy}} = - {y}^{- 2} - 2 {y}^{- 3}$

and $g \left(y\right) = 5 {y}^{- 3} + 9 {y}^{- 4}$ and $\frac{\mathrm{dg}}{\mathrm{dy}} = - 15 {y}^{- 4} - 36 {y}^{- 5}$

Therefore

$\frac{\mathrm{dp} \left(y\right)}{\mathrm{dy}} = \left(- {y}^{- 2} - 2 {y}^{- 3}\right) \times \left(5 {y}^{- 3} + 9 {y}^{- 4}\right) + \left({y}^{- 1} + {y}^{- 2}\right) \times \left(- 15 {y}^{- 4} - 36 {y}^{- 5}\right)$

= $- 5 {y}^{- 5} - 9 {y}^{- 6} - 10 {y}^{- 6} - 18 {y}^{- 7} - 15 {y}^{- 5} - 36 {y}^{- 6} - 15 {y}^{- 6} - 36 {y}^{- 7}$

= $- 20 {y}^{- 5} - 70 {y}^{- 6} - 54 {y}^{- 7}$