Question

If `y=ln(2x^2-6x)`, then what is `dy/dx`?

Answer 1

The answer is `=(2x-3)/(x(x-3))`

Explanation:

This is a chain differentiation

`y=ln u(x)`

`dy/dx=1/(u(x))*u'(x)`

`y=ln(2x^2-6x)`

`dy/dx=1/(2x^2-6x)*(4x-6)`

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