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प्रश्न
Differentiate \[\frac{e^x \log x}{x^2}\] ?
उत्तर
\[\text{Let} y = \frac{e^x \log x}{x^2}\]
\[\text{Differentiate with respect to x we get}, \]
\[\frac{d y}{d x} = \frac{x^2 \frac{d}{dx}\left( e^x \log x \right) - \left( e^x \log x \right)\frac{d}{dx} x^2}{\left( x^2 \right)^2} \left[ \text{Using quotient rule} \right]\]
\[ = \frac{x^2 \left\{ e^x \frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( e^x \right) \right\} - e^x \log x \times 2x}{x^4} \left[ \text{Using product rule} \right]\]
\[ = \frac{x^2 \left[ \frac{e^x}{x} + e^x \log x \right] - 2x e^x \log x}{x^4}\]
\[ = \frac{\frac{x^2 e^x \left( 1 + x\log x \right)}{x} - 2x e^x \log x}{x^4}\]
\[ = \frac{x e^x \left[ 1 + x\log x - 2\log x \right]}{x^4}\]
\[ = \frac{x e^x}{x^3}\left[ \frac{1}{x} + \frac{x \log x}{x} - \frac{2\log x}{x} \right]\]
\[ = e^x x^{- 2} \left[ \frac{1}{x} + \log x - \frac{2}{x}\log x \right]\]
\[So, \frac{d}{dx}\left[ \frac{e^x \log x}{x^2} \right] = e^x x^{- 2} \left[ \frac{1}{x} + \log x - \frac{2}{x}\log x \right]\]
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