Question

The derivative of ${\tan }^{-1}\frac{2x}{1-x^2}$ with respect to ${\sin }^{-1}\frac{2x}{1+x^2}$, is

• A. $1\text{for all x}$
• B. $1\text{for}|x|>1\text{and}-1\text{for}|x|<1$
• C. $1\text{for}|x|<1\text{and}-1\text{for}|x|>1$
• D. $1\text{for}|x|\le 1\text{and}-1\text{for}|x|>1$

Answer 1

Answer: (C)

$1\text{for}|x|<1\text{and}-1\text{for}|x|>1$

Let $y={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ and
$z={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$y=\{\begin{array}{ccc}2{\tan }^{-1}x& & ,\text{if}-1<x<1\\ 2{\tan }^{-1}x-\pi & & ,\text{if}x>1\\ 2{\tan }^{-1}x+\pi & & ,\text{if}x<-1\end{array}$
$z=\{\begin{array}{ccc}2{\tan }^{-1}x& & ,\text{if}-1\le x\le 1\\ \pi -2{\tan }^{-1}x& & ,\text{if}x>1\\ -\pi -2{\tan }^{-1}x& & ,\text{if}x<-1\end{array}$
$\therefore \frac{dy}{dz}=\{\begin{array}{ccc}1& & \text{, if}-1<x<1\\ -1& & \text{, if}|x|>1\end{array}$