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

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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 for question: Differentiate √ Tan − 1 ( X 2 ) ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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