The differential coefficient of tan^{-1}sqrt{ | vedclass

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(D) Let $y_1 = 	an^{-1}\sqrt{x}$ and $y_2 = \sqrt{x}$.Differentiating $y_1$ with respect to $x$ using the chain rule:$\frac{dy_1}{dx} = \frac{1}{1+(\

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$\frac{1}{1+x}$

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(D) Let $y_1 = 	an^{-1}\sqrt{x}$ and $y_2 = \sqrt{x}$.Differentiating $y_1$ with respect to $x$ using the chain rule:$\frac{dy_1}{dx} = \frac{1}{1+(\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x}) = \frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}$.Differentiating $y_2$ with respect to $x$:$\frac{dy_2}{dx} = \frac{1}{2\sqrt{x}}$.Now,the differential coefficient of $y_1$ with respect to $y_2$ is given by:$\frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} = \frac{\frac{1}{1+x} \cdot \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} = \frac{1}{1+x}$.Thus,the correct option is $D$.

The differential coefficient of $\tan^{-1}\sqrt{x}$ with respect to $\sqrt{x}$ is

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