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प्रश्न
पर्याय
\[\frac{x^2 - 1}{x^2 - 4}\]
1
\[\frac{x^2 + 1}{x^2 - 4}\]
\[e^x \frac{x^2 - 1}{x^2 - 4}\]
उत्तर
\[\frac{x^2 - 1}{x^2 - 4}\]
\[\text { Let y } = \frac{d}{dx}\left[ \log\left\{ e^x \left( \frac{x - 2}{x + 2} \right)^\frac{3}{4} \right\} \right]\]
\[ \Rightarrow y = \frac{d}{dx}\left[ x\log e + \frac{3}{4}\log\left( \frac{x - 2}{x + 2} \right) \right]\]
\[ \Rightarrow y = \frac{d}{dx}\left[ x + \frac{3}{4}\log\left( \frac{x - 2}{x + 2} \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = 1 + \frac{3}{4\left( \frac{x - 2}{x + 2} \right)} \times \frac{\left( x + 2 \right) \times 1 - \left( x - 2 \right) \times 1}{\left( x + 2 \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = 1 + \frac{3\left( x + 2 \right)}{4\left( x - 2 \right)} \times \frac{x + 2 - x + 2}{\left( x + 2 \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = 1 + \frac{3\left( x + 2 \right)}{4\left( x - 2 \right)} \times \frac{4}{\left( x + 2 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = 1 + \frac{3}{\left( x^2 - 4 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x^2 - 4 + 3}{x^2 - 4}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x^2 - 1}{x^2 - 4}\]
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