# What is derivative of ( cos x )^sinx ?

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### 1 Answer

The given composed exponent function requests to you to use the logarithmic differentiation method, such that:

`y = cos x^sin x => ln y = ln (cos x)^sin x`

Using the power property of logarithms, yields:

`ln y = sin x*ln (cos x)`

Differentiating both sides, using the chain rule and product rule, yields:

`(1/y)*y' = (sin x)'*ln cos x + sin x*(ln cos x)'`

`(1/y)*y' = cos x*ln cos x + sin x*1/(cos x)*(-sin x)`

`y' = y*(cos x*ln cos x - (sin^2 x)/(cos x))`

Replacing `(cos x)^sin x` for `y` yields:

`y' = (cos x)^sin x*(ln (cos x)^cos x - (sin^2 x)/(cos x))`

**Hence, evaluating the derivative of the given function, using logarithmic differentiation, yields **`y' = (cos x)^sin x*(ln (cos x)^cos x - (sin^2 x)/(cos x)).`

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