Question

How do you differentiate `f(x)= ln(x^2-3x)`?

Answer 1

`f' (x)=(2x-3)/(x^2-3x)`

Explanation:

The formula for derivative of `ln u` is

`d/dx(ln u)=(d/dxu)/ln u`

`f' (x)=d/dx(f (x))=(1/ln (x^2-3x))*d/dx(x^2-3x)`

`f' (x)=(1/ln (x^2-3x))(2x-3)`

`f' (x)=(2x-3)/(x^2-3x)`

God bless...I hope the explanation is useful.