The differential coefficient of sin^{-1}left( | vedclass

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(D) Let $y = \sin^{-1}\left(\frac{1-x}{1+x}\right)$.To differentiate $y$ with respect to $x$,we use the chain rule: $\frac{dy}{dx} = \frac{1}{\sqrt{1

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None of these

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(D) Let $y = \sin^{-1}\left(\frac{1-x}{1+x}\right)$.To differentiate $y$ with respect to $x$,we use the chain rule: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - (\frac{1-x}{1+x})^2}} \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right)$.$\frac{dy}{dx} = \frac{1}{\sqrt{\frac{(1+x)^2 - (1-x)^2}{(1+x)^2}}} \cdot \frac{-(1+x) - (1-x)}{(1+x)^2} = \frac{1+x}{\sqrt{4x}} \cdot \frac{-2}{(1+x)^2} = \frac{-2}{2\sqrt{x}(1+x)} = \frac{-1}{\sqrt{x}(1+x)}$.Now,let $z = \sqrt{x}$. Then $\frac{dz}{dx} = \frac{1}{2\sqrt{x}}$.We need to find $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-1}{\sqrt{x}(1+x)} \div \frac{1}{2\sqrt{x}} = \frac{-1}{\sqrt{x}(1+x)} \cdot 2\sqrt{x} = \frac{-2}{1+x}$.Since this result is not among the options $A$,$B$,or $C$,the correct option is $D$.

The differential coefficient of $\sin^{-1}\left(\frac{1-x}{1+x}\right)$ with respect to $\sqrt{x}$ is:

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