How do you differentiate g(y) =(x^2 + 6) (4x^6 + 5) using the product rule?

Jan 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \left(15 {x}^{6} + 72 {x}^{4} + 5\right)$

Explanation:

Product rule:

$f ' \left(u v\right) = u v ' + v u '$

$g \left(y\right) = \left({x}^{2} + 6\right) \left(4 {x}^{6} + 5\right)$

$g ' \left(y\right) = 2 x \left(4 {x}^{6} + 5\right) + 24 {x}^{5} \left({x}^{2} + 6\right)$

$= 8 {x}^{7} + 10 x + 24 {x}^{7} + 144 {x}^{5}$

$= 30 {x}^{7} + 144 {x}^{5} + 10 x$

$= 2 x \left(15 {x}^{6} + 72 {x}^{4} + 5\right)$