# How do you find the derivative of p(y) = (y^-1 + y^-2)(5y^-3 + 9y^-4)?

Jun 26, 2015

I found: $p ' \left(y\right) = - 54 {y}^{-} 7 - 70 {y}^{-} 6 - 20 {y}^{-} 5$
$p ' \left(y\right) = \left(- {y}^{-} 2 - 2 {y}^{-} 3\right) \left(5 {y}^{-} 3 + 9 {y}^{-} 4\right) + \left({y}^{-} 1 + {y}^{-} 2\right) \left(- 15 {y}^{-} 4 - 36 {y}^{-} 5\right) =$
$= - 5 {y}^{-} 5 - 9 {y}^{-} 6 - 10 {y}^{-} 6 - 18 {y}^{-} 7 - 15 {y}^{-} 5 - 36 {y}^{-} 6 - 15 {y}^{-} 6 - 36 {y}^{-} 7 =$
$= - 54 {y}^{-} 7 - 70 {y}^{-} 6 - 20 {y}^{-} 5$