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