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