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प्रश्न
If \[y = \log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}\] ,show that \[\frac{dy}{dx} = \frac{- 1}{2\sqrt{x^2 - 1}}\] ?
उत्तर
\[\text{Here} , y = \log\left( \sqrt{x - 1} - \sqrt{x + 1} \right)\]
Differentiate it with respect to x we get,
\[\frac{d y}{d x} = \frac{d}{dx}\log\left( \sqrt{x - 1} - \sqrt{x + 1} \right)\]
\[ = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\frac{d}{dx}\left( \sqrt{x - 1} - \sqrt{x + 1} \right) \left[ \text{Using chain rule} \right]\]
\[ = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left[ \frac{d}{dx}\sqrt{x - 1} - \frac{d}{dx}\sqrt{x + 1} \right]\]
\[ = \frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left[ \frac{1}{2} \left( x - 1 \right)^\frac{- 1}{2} - \frac{1}{2} \left( x + 1 \right)^\frac{- 1}{2} \right]\]
\[ = \frac{1}{2}\frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left( \frac{1}{\sqrt{x - 1}} - \frac{1}{\sqrt{x + 1}} \right)\]
\[ = \frac{1}{2}\frac{1}{\left( \sqrt{x - 1} - \sqrt{x + 1} \right)}\left\{ \frac{- \left( \sqrt{x - 1} - \sqrt{x + 1} \right)}{\left( \sqrt{x - 1} \right)\left( \sqrt{x + 1} \right)} \right\}\]
\[ = \frac{- 1}{2}\left( \frac{1}{\left( \sqrt{x - 1} \right)\left( \sqrt{x + 1} \right)} \right)\]
\[ = \frac{- 1}{2\sqrt{x^2 - 1}}\]
\[So, \frac{d y}{d x} = \frac{- 1}{2\sqrt{x^2 - 1}}\]
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