Question

How do you find the derivative of `(ln(sin(2x)))^2`?

Answer 1

`f'(x)=4ln(sin(2x))*cot(2x)`

Explanation:

Chain rule!

`f(x)=(ln(sin(2x)))^2`

let `u=ln(sin(2x))`
now `f(x)=u^2`

so `f'(x)=2u*(du)/dx = 2ln(sin(2x))*(du)/dx`

now to find `(du)/dx`

let `w=sin(2x)`
now `u=ln(w)`

so `(du)/dx=1/w*(dw)/(dx) = 1/sin(2x)*(dw)/(dx)`

now to find `(dw)/(dx)`

let `s=2x`
now `w=sin(s)`

so `(dw)/(dx)=cos(s)*(ds)/(dx) = cos(s)*(ds)/(dx)`

We know that `(ds)/dx = 2` so...

`(dw)/(dx)=cos(2x)*2`

`(du)/dx= 1/sin(2x)*cos(2x)*2`

`f'(x)=2ln(sin(2x))*1/sin(2x)*cos(2x)*2`

simplify

`f'(x)=4ln(sin(2x))*cos(2x)/sin(2x)`

`f'(x)=4ln(sin(2x))*cot(2x)`