Question

find the derivative of the function: y=ln(2x^3-x^2) can some please show me how to do the work so i can try the other questions by myself thanks

Answer 1

\[\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}\] by the chain rule

Answer 2

Okay, there are 3 things you'll use here. The first and most basic is (x^n)' = nx^(n-1)
The second is that (ln(x))' = 1/x
the third is chain rule. Chain rule tells us how to do derivatives of functions inside of functions. In this case, we have ln(2x^3 - x^2). (2x^3-x^2) is inside of the function ln(x).
Chain rule is: f(g(x))' = f'(g(x))*g'(x), that is, the derivative of the outside function multiplied by the derivative of the inside function.
Here's how it works out in this case:
f(x) = ln(2x^3-x^2)
f'(x) = 1/(2x^3-x^2) * (6x^2 - 2x)
The first part is the derivative of ln. The second part is the derivative of the inside function, 2x^3-x^2.

Answer 3

Please ask me any questions you have.