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प्रश्न
Find \[\frac{dy}{dx}\]
\[y = x^x + x^{1/x}\] ?
उत्तर
\[\text{ We have, y } = x^x + x^\frac{1}{x} \]
\[ \Rightarrow y = e^{\log x^x }+ e^{\log x^{\frac{1}{x}} }\]
\[ \Rightarrow y = e^{x \log x} + e^\left( \frac{1}{x}\log x \right)\]
Differentiating with respect to x using chain rule,
\[\frac{dy}{dx} = \frac{d}{dx}\left( e^{x \log x} \right) + \frac{d}{dx}\left( e^{\frac{1}{x}\log x }\right)\]
\[ = e^{x \log x} \frac{d}{dx}\left( x \log x \right) + e^{\frac{1}{x}\log x} \frac{d}{dx}\left( \frac{1}{x}\log x \right)\]
\[ = e^{\log x^x} \left[ x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x \right) \right] + e^{ \log x^{\frac{1}{x}}} \left[ \frac{1}{x}\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( \frac{1}{x} \right) \right] \]
\[ = \left( x \right)^x \left[ x\left( \frac{1}{x} \right) + \log x\left( 1 \right) \right] + x^\frac{1}{x} \left[ \left( \frac{1}{x} \right)\left( \frac{1}{x} \right) + \log x\left( - \frac{1}{x^2} \right) \right]\]
\[ = \left( x \right)^x \left[ 1 + \log x \right] + x^\frac{1}{x} \left( \frac{1}{x^2} - \frac{1}{x^2}\log x \right)\]
\[ = x^x \left[ 1 + \log x \right] + x^\frac{1}{x} \left( \frac{1 - \log x}{x^2} \right)\]
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