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

1 Answer
Jul 11, 2016

#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))#