Question

What is the derivative of `y = ln(2x^3-x^2)`?

Answer 1

`d/(dx)lnf(x)=(f'(x))/f(x)`

Explanation:

Recall that `d/(dx)lnx=1/x`

`y=ln(2x^3-x^2)` is a function within a function; we can apply chain rule.

Let `u=2x^3-x^2`, then `y=lnu`
`(dy)/(du)=1/u` and `(du)/(dx)=6x^2-2x`
`(dy)/(dx)=(dy)/(du)`x`(du)/(dx)=1/(2x^3-x^2)*(6x^2-2x)=(6x^2-2x)/(2x^3-x^2)`