# Find the derivative of the function y=x² /arctgx ?

##### 1 Answer
Jun 14, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\arctan x\right)}^{-} 2 \left(2 x \arctan x - {x}^{2} / \left({x}^{2} + 1\right)\right)$

#### Explanation:

$y = {x}^{2} / \arctan x$

$y = {x}^{2} {\left(\arctan x\right)}^{-} 1$

Let's differentiate this using the product rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{d}{\mathrm{dx}} {x}^{2}\right) {\left(\arctan x\right)}^{-} 1 + {x}^{2} \left(\frac{d}{\mathrm{dx}} {\left(\arctan x\right)}^{-} 1\right)$

The derivative of ${x}^{2}$ can be found with the product rule. To find the derivative of ${\left(\arctan x\right)}^{-} 1$, use the chain rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {\left(\arctan x\right)}^{-} 1 + {x}^{2} \left(- {\left(\arctan x\right)}^{-} 2\right) \left(\frac{d}{\mathrm{dx}} \arctan x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {\left(\arctan x\right)}^{-} 1 - {x}^{2} {\left(\arctan x\right)}^{-} 2 \left(\frac{1}{{x}^{2} + 1}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\arctan x\right)}^{-} 2 \left(2 x \arctan x - {x}^{2} / \left({x}^{2} + 1\right)\right)$