# What is the derivative of y=arctan sqrt((1-x)/(1+x))?

Nov 26, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\sqrt{{\left(1 + x\right)}^{5}}}{2 \sqrt{1 - x}}$

#### Explanation:

Let $u = \sqrt{\frac{1 - x}{1 + x}}$. Also observe $\frac{1 - x}{1 + x} = 1 - \frac{2 x}{1 + x}$

then using chain rule $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\frac{1 - x}{1 + x}}} \times \frac{d}{\mathrm{dx}} \left(1 - \frac{2 x}{1 + x}\right)$

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

= $\frac{\sqrt{1 + x}}{2 \sqrt{1 - x}} \times \left(- \frac{2 \left(1 + x\right) - 2 x}{1 + x} ^ 2\right)$

= $\frac{\sqrt{1 + x}}{2 \sqrt{1 - x}} \times \left(- \frac{2}{1 + x} ^ 2\right)$

= $- {\sqrt{1 + x}}^{3} / \sqrt{1 - x}$

Hence $y = \arctan u$ and hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {u}^{2}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + \frac{1 - x}{1 + x}} \times \left(- \frac{\sqrt{{\left(1 + x\right)}^{3}}}{\sqrt{1 - x}}\right)$

= $- \frac{1 + x}{2} \times \frac{\sqrt{{\left(1 + x\right)}^{3}}}{\sqrt{1 - x}}$

= $- \frac{\sqrt{{\left(1 + x\right)}^{5}}}{2 \sqrt{1 - x}}$