Question

What is the derivative of `(cosx)^x`?

Answer 1

`d/dx((cosx)^x) = (lncosx -x/sinx)(cosx)^x`

Explanation:

Let `f(x) =(cosx)^x`

`lnf(x)= xlncosx`

`(f'(x))/(f(x)) = lncosx - x/sinx`

`f'(x) = f(x)(lncosx-x/sinx)`

`f'(x) = (lncosx -x/sinx)(cosx)^x`

Answer 2

`d/dxcos^x(x)=cos^x(x)(-xtan(x)+ln(cos(x)))`

Explanation:

Step 1. Express `cos(x)^x` as a power of `e`.

`d/dxcos^x(x)=d/dxe^(ln(cos^x(x)))=d/dxe^(xln(cos(x)))`

Step 2. Use the chain rule with `u=xln(cos(x))` and `d/(du)(e^u)=e^u`

Chain Rule: `d/dx(e^(xln(cos(x))))=(de^u)/(du)(du)/(dx)`

`(de^u)/(du)=e^u`

`(du)/(dx)=d/dxxln(cos(x))`

Using the Product Rule

`(du)/(dx)=d/dxxln(cos(x))=x d/dxln(cos(x))+ln(cos(x))dx/dx`

`(du)/(dx)=x(1/cos(x)(-sin(x)))+ln(cos(x))(1)`

`(du)/(dx)=-xtan(x)+ln(cos(x))`

Plugging it back in to the Chain Rule

`color(blue)((de^u)/(du))color(red)((du)/(dx))=color(blue)(e^(xln(cos(x))))color(red)((-xtan(x)+ln(cos(x))))`

Simplifying a bit gives

`(de^u)/(du)(du)/(dx)=cos^x(x)(-xtan(x)+ln(cos(x)))`