Question

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

Answer 1

Chain rule: `(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dw)(dw)/(dx)`

Explanation:

To do so, we'll rename `f(x)=ln(u)`, then `u=sqrt(v)`, `v=sin(w)` and finally `w=2x`

`(dy)/(dx)=1/u(1/(2sqrt(v)))cos(w)(2)=(cancel(2)cos(w))/(cancel(2)usqrt(v))`

Substituting `u`, `v` and `w`, in this order:

`(dy)/(dx)=(cos(w))/(sqrt(v)sqrt(v))=(cos(w))/v=(cos(w))/sin(w)=cos(2x)/sin(2x)=color(green)(cot(2x))`