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Question
If \[y = \log \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]prove that \[\frac{dy}{dx} = \frac{x - 1}{2x \left( x + 1 \right)}\] ?
Solution
\[\text{ We have, y } = \log\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]
Differentiate it with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)\]
\[ = \frac{1}{\sqrt{x} + \frac{1}{\sqrt{x}}}\frac{d}{dx}\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \]
\[ = \frac{\sqrt{x}}{x + 1}\left( \frac{1}{2\sqrt{x}} - \frac{1}{2x\sqrt{x}} \right)\]
\[ = \frac{1}{2}\frac{\sqrt{x}}{x + 1}\left( \frac{x - 1}{x\sqrt{x}} \right)\]
\[ = \frac{x - 1}{2x\left( x + 1 \right)}\]
\[So, \frac{d y}{d x} = \frac{x - 1}{2x\left( x + 1 \right)}\]
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