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If Y = Log ( √ X + 1 √ X ) Prove that D Y D X = X − 1 2 X ( X + 1 ) ? - CBSE (Commerce) Class 12 - Mathematics

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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

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

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

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

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

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

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

  Is there an error in this question or solution?

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Solution If Y = Log ( √ X + 1 √ X ) Prove that D Y D X = X − 1 2 X ( X + 1 ) ? Concept: Simple Problems on Applications of Derivatives.
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