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