How do you differentiate $f (x) = ln (x^{2} - 3 x)$ $f(x)= ln(x^2-3x)$ ?

Question

3933 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-04T14:22:39

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

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.

How do you differentiate f(x)= ln(x^2-3x)f(x)=ln(x2−3x)f(x)= ln(x^2-3x)?