# How do you find the derivative of sqrt[arctan(x)]?

Sep 21, 2016

$\frac{1}{2 \left({x}^{2} + 1\right) \sqrt{\arctan} \left(x\right)}$

#### Explanation:

$y = \sqrt{\arctan} \left(x\right)$

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

We will use the chain rule. The specific adaptation of the chain rule that we will be using here is where we use the power rule on the outside function:

$\frac{d}{\mathrm{dx}} {u}^{\frac{1}{2}} = \frac{1}{2} {u}^{- \frac{1}{2}} \frac{\mathrm{du}}{\mathrm{dx}}$

So:

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

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

Note that $\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{{x}^{2} + 1}$:

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

Sep 21, 2016

$\frac{1}{2 \left({x}^{2} + 1\right) \sqrt{\arctan} \left(x\right)}$

#### Explanation:

$y = \sqrt{\arctan} \left(x\right)$

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

Differentiate both sides. The chain rule will be used on the left!

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{x}^{2} + 1}$

Since $2 y = 2 \sqrt{\arctan} \left(x\right)$:

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

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