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

1 Answer
Jan 3, 2016

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))