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# Solution for Differentiate √ Tan − 1 ( X 2 ) ? - CBSE (Commerce) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

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

#### Solution

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

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

Differentiate it with respect to x we get,

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

$= \frac{1}{2} \left\{ \tan^{- 1} \left( \frac{x}{2} \right) \right\}^\frac{1}{2} - 1 \frac{d}{dx}\left( \tan^{- 1} \frac{x}{2} \right) \left[ \text{Using chain rule }\right]$

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

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

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

$So, \frac{d}{dx}\left\{ \sqrt{\tan^{- 1} \left( \frac{x}{2} \right)} \right\} = \frac{1}{\left( 4 + x^2 \right)\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}}$

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Solution Differentiate √ Tan − 1 ( X 2 ) ? Concept: Simple Problems on Applications of Derivatives.
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