Question

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

Answer 1

It'll take chain rule to find the solution.

Explanation:

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

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

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

Substituting `u`, then `v`:

`(df(x))/(dx)=((2x-1)cosv)/sinv=((2x-1)cos(x^2-x))/(sin(x^2-x))`

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