Continuity and Differentiability — Class 12 Maths Solution

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

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

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

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

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

$\Rightarrow \frac{dy}{dx} = \frac{{ - 1}}{{2\sqrt x }} \cdot \left( {\frac{1}{{x + 1}}} \right)$