Differentiate `(cosx)^(sinx)` with respect to `(sinx)^(cosx)`

Explanation

Let. `u=(cos x)^{sin x}`


Taking log on both sides,



`log u=log (cos x)^{sin x} `


`Rightarrow log u=sin x log (cos x)`



Differentiating it with respect to `x` using chain rule,


`frac{1}{u} frac{d u}{d x}=sin x frac{d}{d x}(log cos x)+log cos x frac{d}{d x}(sin x)` [using product rule ]


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