Question

If $y={\sec }^{-1}\frac{2x}{1+x^2}+{\sin }^{-1}\frac{x-1}{x+1}$ ,then $\frac{dy}{dx}$ is equal to :

• A. 1
• B. $\frac{x-1}{x+1}$
• C. does not exist
• D. none of these

Answer 1

Answer: (C)

Let $y={\sec }^{-1}\left(\frac{2x}{1+x^2}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)$
Put $x=\tan \theta$ , we get
$\frac{2x}{1+x^2}=\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$
Since, $-1\le \sin \theta \le 1$
$\therefore -1\le \frac{2x}{1+x^2}\le 1$
$\Rightarrow {\sec }^{-1}\left(\frac{2x}{1+x^2}\right)$ is defined only at
$x=1,-1$
$\therefore \frac{dy}{dx}$ does not exist.