Question

How do you differentiate `(cos x)^(sin x)`?

Answer 1

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

Explanation:

This problem requires us to make use of logarithmic differentiation, as well as the product rule for derivatives.

Let `y = (cosx)^(sinx)`

Taking the natural logarithm of both sides yields

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

Now we have to take a derivative on each side of the equation.

Remembering that

`d/dx[ln(u)] = (u')/(u)` and `d/dx[f(x)g(x)] = f(x)g'(x)+f'(x)g(x)`

We now get

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

Isolating our `y'`-term gives us

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

Substituting `y` back into the equation gives us our final result of

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