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

1 Answer
Dec 25, 2015

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