# How do you find the derivative of arctan [(1-x)/(1+x)]^(1/2)?

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) = - \frac{1}{2} \cdot \frac{\sqrt{1 - {x}^{2}}}{\left(1 - {x}^{2}\right)}$

#### Explanation:

Formula for finding derivative of arctangent

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 u\right) = \left(\frac{1}{1 + {u}^{2}}\right) \frac{d}{\mathrm{dx}} \left(u\right)$

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

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) =$
$\left(\frac{1}{1 + \frac{1 - x}{1 + x}}\right) \left(\frac{1}{2 \left(\sqrt{\frac{1 - x}{1 + x}}\right)}\right) \cdot \frac{d}{\mathrm{dx}} \left(\frac{1 - x}{1 + x}\right)$

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) =$
$\left(\frac{1}{1 + \frac{1 - x}{1 + x}}\right) \left(\frac{1}{2 \left(\sqrt{\frac{1 - x}{1 + x}}\right)}\right) \cdot \left(\frac{\left(1 + x\right) \left(- 1\right) - \left(1 - x\right) \left(1\right)}{1 + x} ^ 2\right)$

Simplify

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) =$
$\left(\frac{1 + x}{1 + x + 1 - x}\right) \left(\frac{1}{2} \cdot \sqrt{\frac{1 + x}{1 - x}}\right) \cdot \left(\frac{- 1 - x - 1 + x}{1 + x} ^ 2\right)$

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) =$
$\left(\frac{1 + x}{2}\right) \left(\frac{1}{2} \cdot \sqrt{\frac{1 + x}{1 - x}}\right) \cdot \left(\frac{- 2}{1 + x} ^ 2\right)$

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) =$
$\left(\frac{1}{2} \cdot \sqrt{\frac{1 + x}{1 - x}}\right) \cdot \left(\frac{- 1}{1 + x}\right)$

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) =$
$\left(\frac{1}{2} \cdot \sqrt{\left(\frac{1 + x}{1 - x}\right) \left(\frac{1 - x}{1 - x}\right)}\right) \cdot \left(\frac{- 1}{1 + x}\right)$

$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \sqrt{\frac{1 - x}{1 + x}}\right) = - \frac{1}{2} \cdot \frac{\sqrt{1 - {x}^{2}}}{\left(1 - {x}^{2}\right)}$

God bless....I hope the explanation is useful.