How do you differentiate f(x)= ln(sin(x^2-x)) f(x)=ln(sin(x2x))?

1 Answer
Jan 9, 2016

It'll take chain rule to find the solution.

Explanation:

  • Chain rule: (dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)dydx=dydududvdvdx

In this case, u=sin(v)u=sin(v), v=x^2-xv=x2x, so that we get f(x)=ln(u)f(x)=ln(u) and can now proceed, following the rule.

(df(x))/(dx)=1/ucosv(2x-1)df(x)dx=1ucosv(2x1)

Substituting uu, then vv:

(df(x))/(dx)=((2x-1)cosv)/sinv=((2x-1)cos(x^2-x))/(sin(x^2-x))df(x)dx=(2x1)cosvsinv=(2x1)cos(x2x)sin(x2x)

(df(x))/(dx)=color(green)((2x-1)/tan(x^2-x))df(x)dx=2x1tan(x2x) or (df(x))/(dx)=color(green)((2x-1)cot(x^2-x))df(x)dx=(2x1)cot(x2x)