# How do you find the derivative of y= cos(x)/x^8?

##### 1 Answer
Feb 13, 2016

$y ' = \frac{- \left(x \sin \left(x\right) + 8 \cos \left(x\right)\right)}{x} ^ 9$

#### Explanation:

Use the quotient rule, which states that, for this problem,

$y ' = \frac{{x}^{8} \frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) - \cos \left(x\right) \frac{d}{\mathrm{dx}} \left({x}^{8}\right)}{{x}^{8}} ^ 2$

The two derivatives are:

$\frac{d}{\mathrm{dx}} \left({x}^{8}\right) = 8 {x}^{7} \text{ "" }$(Through the power rule.)

$\frac{d}{\mathrm{dx}} \left(\cos \left(x\right)\right) = - \sin \left(x\right)$

Plugging these both back in, we see that

$y ' = \frac{- {x}^{8} \sin \left(x\right) - 8 {x}^{7} \cos \left(x\right)}{x} ^ 16$

Factor out a $- {x}^{7}$ term.

$y ' = \frac{- {x}^{7} \left(x \sin \left(x\right) + 8 \cos \left(x\right)\right)}{x} ^ 16$

$y ' = \frac{- \left(x \sin \left(x\right) + 8 \cos \left(x\right)\right)}{x} ^ 9$