Question

If \[y = \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]prove that \[\frac{dy}{dx} = \frac{x - 1}{2x \left( x + 1 \right)}\] ?

 

Answer 1

\[\text{ We have, y } = \log\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]

Differentiate it with respect to x, 

\[\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]

\[ = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}}\frac{d}{dx}\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \]

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

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

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

\[So, \frac{d y}{d x} = \frac{x - 1}{2x\left( x + 1 \right)}\]