Question

If $y={\sec }^{-1}\left(\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)+{\sin }^{-1}\left(\frac{x+\sqrt{x}}{\sqrt{x}-1}\right),$ then $\frac{dy}{dx}=$....

• A. x
• B. $\frac{1}{x}$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

Let $y={\sec }^{-1}\left(\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)+{\sin }^{-1}\left(\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)$

$={\cos }^{-1}\left(\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)+{\sin }^{-1}\left(\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\because {\sec }^{-1}\left(x\right)={\cos }^{-1}\left(\frac{1}{x}\right)$

$\therefore y=\frac{\pi }{2}$

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi }{2}\right)=0$