# How do I find the derivative of k(a)=sin^5acos^4a?

## How do I even find f'x for sin^5a and cos^4a? I would use the product rule, but I'm lost on just what the derivatives of the separate parts are.

Apr 30, 2018

$\frac{\mathrm{dk}}{\mathrm{da}} = - 4 {\cos}^{3} a \cdot {\sin}^{6} a + 5 {\sin}^{4} a \cdot {\cos}^{5} a$

#### Explanation:

• Let $g \left(a\right) = {\sin}^{5} a$

g'(a)=5sin^4a*color(blue)(cosa$\textcolor{g r e e n}{\rightarrow \text{Chain Rule}}$

• Let $h \left(a\right) = {\cos}^{4} a$

$h ' \left(a\right) = 4 {\cos}^{3} a \cdot \left(\textcolor{b l u e}{- \sin a}\right)$color(green)(rarr"Chain Rule")

$k = h \left(a\right) \cdot g \left(a\right)$

$\frac{\mathrm{dk}}{\mathrm{da}} = h ' \left(a\right) \cdot g \left(a\right) + g ' \left(a\right) \cdot h \left(a\right)$

Substitute

$\frac{\mathrm{dk}}{\mathrm{da}} = \left(- 4 {\cos}^{3} a \cdot \sin a\right) \cdot \left({\sin}^{5} a\right) + \left(5 {\sin}^{4} a \cdot \cos a\right) \cdot \left({\cos}^{4} a\right)$

$\frac{\mathrm{dk}}{\mathrm{da}} = - 4 {\cos}^{3} a \cdot {\sin}^{6} a + 5 {\sin}^{4} a \cdot {\cos}^{5} a$