Question

If $y={\sec }^{-1}\left(\frac{x+1}{x-1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)$ then $\frac{dy}{dx}$ is equal to

• A. $0$
• B. $x+1$
• C. $1$
• D. $-1$

Answer 1

Answer: (A)

$0$

$y={\sec }^{-1}\left(\frac{x+1}{x-1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)$
$={\cos }^{-1}\left(\frac{x-1}{x+1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)=\frac{\pi }{2}$
However the above function is defined only for values of x, given by
$-1\le \frac{x-1}{x+1}\le 1$
i.e. $\frac{x-1}{x+1}+1\ge 0$ and $\frac{x-1}{x+1}-1\le 0$
i.e. $\frac{2x}{x+1}\ge 0$ and $\frac{2}{x+1}\ge 0$
i.e $x<-1$ or $\ge 0$ and $x>-1$
i.e $x\ge 0$
Hence we have
$y=\frac{\pi }{2},x\ge 0\Rightarrow \frac{dy}{dx}=0,x>0$