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

0

The correct answer is Option (1) → 0

Given:

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

Observe that:

$\sqrt{ \frac{x}{x+1} } = \cos \theta \quad \Rightarrow \quad \sqrt{ \frac{x+1}{x} } = \sec \theta$

So, define $\theta = \cos^{-1} \left( \sqrt{ \frac{x}{x+1} } \right)$

Then:

$y = \sin^{-1}(\cos \theta) + \sec^{-1}(\sec \theta) = (\frac{\pi}{2} - \theta) + \theta = \frac{\pi}{2}$

Hence, $y$ is constant.

$\Rightarrow \frac{dy}{dx} = 0$