Question

Derivative of (x^3-3x^2+4)/x^2

Answer 1

\[1-8/x^3\]

Answer 2

How did you get the answer? I'm stuck.

Answer 3

Well... i used wolfram alpha to compute it but Im guessing you'd like some working, give me a sec :)

Answer 4

Recall the quotient rule\[dy/dx=(f \prime(x)g(x)-f(x)g \prime(x))/(g(x))^2\]Are you using that formula?

Answer 5

That's a really long and bad way of finding it.

Recall basic algebra

Answer 6

\[\frac{x^3-3x^2+4}{x^2} = \frac{x^3}{x^2}-\frac{3x^2}{x^2} +\frac{4}{x^2} = x -3 +4x^{-2}\]

Answer 7

So, let y= x -3 +4x^-2

dy/dx= 1 - 8x^-3

\[\frac{dy}{dx}= 1- \frac{8}{x^3}\]