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Question
Differentiate \[\left( \cos x \right)^{\sin x }\] with respect to \[\left( \sin x \right)^{\cos x }\]?
Solution
\[\text { Let, u } = \left( \cos x \right)^{\sin x} \]
Taking log on both sides,
\[\log u = \log \left( \cos x \right)^{\sin x } \]
\[ \Rightarrow \log u = \sin x \log\left( \cos x \right)\]
Differentiating it with respect to x using chain rule,
\[\frac{1}{u}\frac{du}{dx} = \sin x\frac{d}{dx}\left( \log \cos x \right) + \log \cos x\frac{d}{dx}\left( \sin x \right) \left[ \text{ using product rule } \right]\]
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = \sin x\left( \frac{1}{\cos x} \right)\frac{d}{dx}\left( \cos x \right) + \log \cos x\left( \cos x \right)\]
\[ \Rightarrow \frac{du}{dx} = u\left[ \left( \tan x \right) \times \left( - \sin x \right) + \log \log x\left( \cos x \right) \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( \cos x \right)^{ \sin x } \left[ \cos x \log\cos x - \sin x \tan x \right] . . . \left( i \right)\]
\[\text { Let, v }= \left( \sin x \right)^{\cos x }\]
Taking log on both sides,
\[\log v = \log \left( \sin x \right)^{\cos x} \]
\[ \Rightarrow \log v = \cos x \log\left( \sin x \right)\]
Differentiating it with respect to x using chain rule,
\[\frac{1}{v}\frac{dv}{dx} = \cos x\frac{d}{dx}\left( \log\sin x \right) + \log\sin x\frac{d}{dx}\left( \cos x \right) ..........\left[ \text { using product rule } \right]\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = \cos x\left( \frac{1}{\sin x} \right)\frac{d}{dx}\left( \sin x \right) + \log\sin x\left( - \sin x \right)\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( \sin x \right)^{\cos x } \left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]\]
\[\text { dividing equation }\left( i \right) by \left( ii \right), \]
\[ \therefore \frac{du}{dv} = \frac{\left( \cos x \right)^{\sin x } \left[ \cos x \log\cos x - \sin x \tan x \right]}{\left( \sin x \right)^{\cos x } \left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]}\]
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