Question

What is the derivative of `(ln(sin(2x)))^2`?

Answer 1

`d/dx((ln(sin(2x)))^2)=4cot2xln(sin(2x))`

Explanation:

If we let `y=(ln(sin(2x)))^2`, then we can find the derivative of `y` using the chain rule. But first, we will have to define the composition of the functions we need to differentiate.

Let `y=u^2`

and `u=lnv`

and `v=sinw`

and `w=2x`

Then,

`dy/dx=dy/(du)xx(du)/(dv)xx(dv)/(dw)xx(dw)/dx`

`dy/(du)=2u=2lnv=2ln(sinw)=2ln(sin(2x))`

`(dv)/(du)=1/v=1/sinw=1/sin(2x)`

`(dv)/(dw)=cosw=cos(2x)`

`(dw)/dx=2`

`dy/dx=2ln(sin(2x))xx1/sin(2x)xxcos(2x)xx2`

`=4cot(2x)ln(sin(2x))`