Question

can someone help me find: y' for y=ln(2x^3-3x) ?

Answer 1

The derivative of ln(u) = 1/u (du)

Answer 2

So the derivative will be:

1/(2x^3 - 3x) multiplied by (6x^2 - 3)

Or, (6x^2 - 3)/(2x^3 - 3x) which can be simplified.

Answer 3

this is more of a calculus ii derivative...logarithims