# How do you find the derivative using the power chain rule of y=(cos(x^4)+x^3))^8?

Aug 10, 2015

$8 {\left(\cos {x}^{4} + {x}^{3}\right)}^{7} \cdot \left(3 {x}^{2} - 4 {x}^{3} \sin {x}^{4}\right)$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 8 {\left(\cos {x}^{4} + {x}^{3}\right)}^{7} \frac{d}{\mathrm{dx}} \left(\cos {x}^{4} + {x}^{3}\right)$

= $8 {\left(\cos {x}^{4} + {x}^{3}\right)}^{7} \cdot \left(\frac{d}{\mathrm{dx}} \cos {x}^{4} + \frac{d}{\mathrm{dx}} {x}^{3}\right)$

=$8 {\left(\cos {x}^{4} + {x}^{3}\right)}^{7} \cdot \left(- \sin {x}^{4} 4 {x}^{3} + 3 {x}^{2}\right)$

=$8 {\left(\cos {x}^{4} + {x}^{3}\right)}^{7} \left(3 {x}^{2} - 4 {x}^{3} \sin {x}^{4}\right)$