Question

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

Answer 1

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

Differentiate it with respect to x we get,

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

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

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

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

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

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