Question

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

Answer 1

We'll need the chain rule here, which states that `(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)`

Explanation:

Renaming `u=sin(v)` and `v=lnx`, we can follow the rule. Let's derivate it step-by-step:

`(dy)/(du)=2u`

`(du)/(dv)=cosv`

`(dv)/(dx)=1/x`

Then:

`(dy)/(dx)=2ucosv(1/x)`

Substituting `u`:

`(dy)/(dx)=2sin(v)cos(v)(1/x)`

Substituting `v`:

`(dy)/(dx)=(2sin(lnx)cos(lnx))/x`