Differentiate the function $\sin^{-1} \frac{1}{\sqrt{x + 1}}$ with respect to $x$.

$\frac{-1}{2\sqrt{x}(x+1)}$

The correct answer is Option (2) → $\frac{-1}{2\sqrt{x}(x+1)}$ ##

Let $y = \sin^{-1} \frac{1}{\sqrt{x + 1}}$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx} = \frac{d}{dx} \sin^{-1} \frac{1}{\sqrt{x + 1}}$

$= \frac{1}{\sqrt{1 - \left( \frac{1}{\sqrt{x + 1}} \right)^2}} \cdot \frac{d}{dx} \frac{1}{(x + 1)^{1/2}} \quad \left[ ∵\frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \right]$

$= \frac{1}{\sqrt{\frac{x + 1 - 1}{x + 1}}} \cdot \frac{d}{dx} \cdot (x + 1)^{-1/2} = \sqrt{\frac{x + 1}{x}} \cdot \frac{-1}{2} (x + 1)^{-\frac{1}{2}-1} \cdot \frac{d}{dx} (x + 1)$

$= \frac{(x + 1)^{1/2}}{x^{1/2}} \cdot \left( -\frac{1}{2} \right) (x + 1)^{-3/2} = \frac{-1}{2\sqrt{x}} \cdot \left( \frac{1}{x + 1} \right)$