Find the derivative of $x ^3$ by using the differential quotient

Question

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!

Answer 1

Hint: It is $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ and $$x^2+xx_0+x_0^2$$ tends to $$3x_0^2$$

Answer 2

Polynomial division:

$\small {(x^3-x_0^3)÷(x-x_0)= x^2 +xx_0+ x_0^2}$.

$\small {-(x^3 -x^2x_0)}$

$-------$

$\small {x^2x_0 -x_0^3}$

$\small{-(x^2x_0 -xx_0^2)}$

$-------$

$\small{xx_0^2-x_0^3}$

$\small{-(xx_0^2-x_0^3)}$

$--------$

$\small{0}$