Prove the following
Question:

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

Solution:

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

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=\frac{d}{d x} \sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x+1}}\right)^{2}}} \cdot \frac{d}{d x}\left(\frac{1}{\sqrt{x+1}}\right)$

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

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

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

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

Thus, $\quad \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}(x+1)}$

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