मराठी

Differentiate Log √ X − 1 X + 1 ?

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प्रश्न

Differentiate \[\log \sqrt{\frac{x - 1}{x + 1}}\] ?

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उत्तर

\[\text{Let } y = \log \sqrt{\frac{x - 1}{x + 1}}\]

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

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

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

Differentiate it with respect to x

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

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

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

\[ = \frac{1}{x^2 - 1}\]

\[So, \frac{d y}{d x} = \frac{1}{x^2 - 1}\]

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पाठ 10: Differentiation - Exercise 11.02 [पृष्ठ ३८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 10 Differentiation
Exercise 11.02 | Q 57 | पृष्ठ ३८
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