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प्रश्न
Differentiate \[\left( \cos x \right)^x + \left( \sin x \right)^{1/x}\] ?
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उत्तर
\[\text{ Let y } = \left( \cos x \right)^x + \left( \sin x \right)^\frac{1}{x} \]
\[ \Rightarrow y = e^{ \log \left( \cos x\right)^x} + e^{\log \left( \sin x \right)^\frac{1}{x} } \]
\[ \Rightarrow y = e^{ x\log\left( \cos x \right) } + e^\frac{1}{x}\log\sin x\]
Differentiating with respect to x,
\[\frac{dy}{dx} = \frac{d}{dx}\left( e^{x \log\cos x} \right) + \frac{d}{dx}\left( e^\frac{1}{x}\log \sin x \right)\]
\[ = e^{x \log\cos x} \times \frac{d}{dx}\left( x \log\cos x \right) + e^\frac{1}{x}\log \sin x \frac{d}{dx}\left( \frac{1}{x}\log\sin x \right)\]
\[ = e^{\log \left( \cos x \right)^x }\times \left[ x\frac{d}{dx}\left( \log\cos x \right) + \log\cos x \times \frac{d}{dx}\left( x \right) \right] + e^{\log \left( \sin x \right)^\frac{1}{x} }\times \left[ \frac{1}{x}\frac{d}{dx}\left( \log\sin x \right) + \log\sin x\frac{d}{dx}\left( \frac{1}{x} \right) \right]\]
\[ = \left( \cos x \right)^x \left[ x\left( \frac{1}{\cos x} \right)\frac{d}{dx}\left( \cos x \right) + \log\cos x\left( 1 \right) \right] + \left( \sin \right)^\frac{1}{x} \left[ \frac{1}{x} \times \frac{1}{\sin x} \times \frac{d}{dx}\left( \sin x \right) + \log\sin x\left( - \frac{1}{x^2} \right) \right]\]
\[ = \left( \cos x \right)^x \left[ x\left( \frac{1}{\cos x} \right)\left( - \sin x \right) + \log\cos x \right] + \left( \sin x \right)^\frac{1}{x} \left[ \frac{1}{x} \times \frac{1}{\sin x}\left( \cos x \right) - \frac{1}{x^2}\log\sin x \right]\]
\[ = \left( \cos x \right)^x \left[ \log\cos x - x \tan x \right] + \left( \sin x \right)^\frac{1}{x} \left[ \frac{\cot x}{x} - \frac{1}{x^2}\log\sin x \right]\]
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