# How do you find the derivative of(x^5+x^6-8)^3?

Jul 18, 2018

Below

#### Explanation:

$y = {\left({x}^{5} + {x}^{6} - 8\right)}^{3}$

Let $u = {x}^{5} + {x}^{6} - 8$
then $\frac{\mathrm{du}}{\mathrm{dx}} = 5 {x}^{4} + 6 {x}^{5}$

Since $u = {x}^{5} + {x}^{6} - 8$, then
$y = {u}^{3}$
$\frac{\mathrm{dy}}{\mathrm{du}} = 3 {u}^{2}$

Therefore,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$
=$3 {u}^{2} \times \left(5 {x}^{4} + 6 {x}^{5}\right)$
=$3 {\left({x}^{5} + {x}^{6} - 8\right)}^{2} \left(5 {x}^{4} + 6 {x}^{5}\right)$