#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.2 question 37

Answer:$\frac{1}{\left(4+x^{2}\right) \sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}}$

Hint: You must know about the rules of solving derivative of Inverse trigonometric function.

Given: $\sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}$

Solution:

Let  $y=\sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}$

$y=\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{\frac{1}{2}}$

Differentiate with respect to x,

$\frac{d y}{d x}=\frac{d}{d x}\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{\frac{1}{2}}$

$\frac{d y}{d x}=\frac{1}{2}\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{\frac{1}{2}-1} \frac{d}{d x} \tan ^{-1}\left(\frac{x}{2}\right)$

$\frac{d y}{d x}=\frac{1}{2}\left\{\tan ^{-1}\left(\frac{x}{2}\right)\right\}^{-\frac{1}{2}} \times \frac{1}{1+\left(\frac{x}{2}\right)^{2}} \times \frac{d}{d x}\left(\frac{x}{2}\right)$

\begin{aligned} &\frac{d y}{d x}=\frac{4}{4\left(4+x^{2}\right) \sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}} \\\\ &\frac{d y}{d x}=\frac{1}{\left(4+x^{2}\right) \sqrt{\tan ^{-1}\left(\frac{x}{2}\right)}} \end{aligned}

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