Question
Differentiate the following w.r.t. x:
$\sin^{-1}\Big(\frac{1}{\sqrt{\text{x}+1}}\Big)$

Answer

Let $\text{y}=\sin^{-1}\Big(\frac{1}{\sqrt{\text{x}+1}}\Big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin^{-1}\Big(\frac{1}{\sqrt{\text{x}+1}}\Big)$
$=\frac{1}{\sqrt{-1\Big(\frac{1}{\sqrt{\text{x+1}}}\Big)^2}}\cdot\frac{\text{d}}{\text{dx}}\frac{1}{(\text{x}+1)^{\frac{1}{2}}}$ $\Big[\because\frac{\text{d}}{\text{dx}}(\sin^{-1}\text{x})=\frac{1}{\sqrt{1-\text{x}^2}}\Big]$ 
$=\frac{1}{\sqrt{\frac{\text{x}+1-1}{\text{x}+1}}}\cdot\frac{\text{d}}{\text{dx}}(\text{x+1})^{\frac{-1}{2}}$
$=\sqrt{\frac{\text{x}+1}{\text{x}}}\cdot\frac{-1}{2}(\text{x}+1)^{\frac{1}{2}-1}\cdot\frac{\text{d}}{\text{dx}}(\text{x+1})$
$=\frac{(\text{x}+1)^{\frac{1}{2}}}{\text{x}^{\frac{1}{2}}}\cdot\Big(-\frac{1}{2}\Big)(\text{x}+1)^{-\frac{3}{2}}$
$=\frac{-1}{2\sqrt{\text{x}}}\cdot\Big(\frac{1}{\text{x}+1}\Big)$

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