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

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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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