What is the derivative of #(cos x)^(sin x)#?

1 Answer
Nov 27, 2016

Let #y = (cosx)^sinx#.

Then #lny = ln(cosx)^sinx#.

We apply the laws of logarithms to simplify as follows:

#lny = sinxln(cosx)#

We now use the chain rule to differentiate #ln(cosx)#.

Letting #y= lnu# and #u = cosx#, we have

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx = 1/u xx -sinx#

#dy/dx= -sinx/cosx#

#dy/dx= -tanx#

Differentiating the entire function now with a combination of implicit differentiation and the product rule, we have:

#1/y(dy/dx) = cosx xx ln(cosx) + sinx xx -tanx#

#dy/dx = y(cosxln(cosx) - sin^2x/cosx)#

#dy/dx= (cosx)^sinx(cosxln(cosx) - sin^2x/cosx)#

Hopefully this helps!