Question

What is the derivative of `f(x)=ln(2x^2-4x)/x`?

Answer 1

`f'(x) = ((2x^2)/(x^2 - 2) - ln(2x^2 - 4x))/(x^2)`

Explanation:

Start by using the chain rule to differentiate `y = ln(2x^2 - 4x)`.

Let `y = lnu` and `u = 2x^2 - 4`.

`dy/dx = 1/u xx 4x = (4x)/(2x^2 - 4) = (2(2x))/(2(x^2 - 2)) = (2x)/(x^2 - 2)`

By the quotient rule:

`f'(x) = ((2x)/(x^2 - 2) xx x - 1 xx ln(2x^2 - 4x))/(x)^2`

`f'(x) = ((2x^2)/(x^2 - 2) - ln(2x^2 - 4x))/(x^2)`

Hopefully this helps!