Find the derivative of the function $y = f(x) = x
^3$ by using the
differential quotient
$$f'(x\_0) = \lim\_{x→x\_0}
\frac{f(x) − f(x\_0)}{x − x\_0}.$$

You must use the polynomial division to simplify it.

Ok so by substituting $x^3$, I get $$\lim\_{x→x\_0}
\frac{x^3 − x\_0^3}{x − x\_0}.$$

Then I use polynom. div. to get 
$x^2+x \cdot x\_0+x\_0^2$
but this does not give me the derivative. So it is wrong. I would appreciate some advice.

Edit: Ok so with @coreyman317 's suggestion I believe I have the right answer.
$x^2+x \cdot x\_0+x\_0^2$

I substitute the limit $x\_0$ into the equation and then I get $x\_0^2+x\_0 \cdot x\_0+x\_0^2=3x\_0^2$
Which is the right answer! Thanks guys!