Question

If $y=si{n}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+se{c}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)$, $x>0$, then $\frac{dy}{dx}$ is equal to

• A. $1$
• B. $0$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

Answer: (B)

$y=si{n}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)+se{c}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)$
$=si{n}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$$+co{s}^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$
$=\frac{\pi }{2}$
$\left\{\because si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}\right\}$
$\Rightarrow y=\frac{\pi }{2}$
$\Rightarrow \frac{dy}{dx}=0$