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

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Question

Differentiate \[e^{\tan^{- 1}} \sqrt{x}\] ?

Solution

\[\text{ Let } y = e^{\tan^{- 1}} \sqrt{x} \]

Differentiate it with respect to x we get,

\[\frac{d y}{d x} = \frac{d}{dx}\left( e^{\tan^{- 1}} \sqrt{x} \right)\]

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

\[ = e^{\tan^{- 1}} \sqrt{x} \times \frac{1}{1 + \left( \sqrt{x} \right)^2}\frac{d}{dx}\left( \sqrt{x} \right)\]

\[ = \frac{e^{\tan^{- 1}} \sqrt{x}}{1 + x} \times \frac{1}{2\sqrt{x}}\]

\[ = \frac{e^{\tan^{- 1}} \sqrt{x}}{2\sqrt{x}\left( 1 + x \right)}\]

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

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