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

Aug 9, 2016

 \ y' = - 1/(2 sqrt [(1-x^2)]

Explanation:

$y = \arctan \sqrt{\frac{1 - x}{1 + x}}$

$\tan y = \sqrt{\frac{1 - x}{1 + x}}$

${\tan}^{2} y = \frac{1 - x}{1 + x}$

${\tan}^{2} y = \frac{2}{1 + x} - 1$

differentiating each side

$2 \tan y {\sec}^{2} y \setminus y ' = - \frac{2}{1 + x} ^ 2$

$\tan y \left({\tan}^{2} y + 1\right) \setminus y ' = - \frac{1}{1 + x} ^ 2$

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

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

$\setminus y ' = - \frac{1}{1 + x} ^ 2 \cdot \sqrt{\frac{1 + x}{1 - x}} \cdot \frac{1 + x}{2}$

 \ y' = - 1/(2 sqrt(1+x) * sqrt [(1-x)]

 \ y' = - 1/(2 sqrt (1-x^2)