# What is the derivative of arctan (8/x^2)?

Apr 1, 2016

$\frac{- 16 x}{{x}^{4} + 64}$

#### Explanation:

By the chain rule:

$y = \arctan u , u = \frac{8}{x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{1 + {u}^{2}} , \frac{\mathrm{du}}{\mathrm{dx}} = - \frac{16}{x} ^ 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$= \frac{1}{1 + {\left(\frac{8}{x} ^ 2\right)}^{2}} \cdot - \frac{16}{x} ^ 3$

$= - \frac{16}{{x}^{3} + \frac{64}{x}} = \frac{- 16 x}{{x}^{4} + 64}$