Question

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

Answer 1

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!