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

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!

• Well first, if you use polynomial long division correctly, your result will be $x^2+xx_0+x_0^2$ instead of $x^2-xx_0+x_0^2$. Now since this is a polynomial in two variables (hence always continuous) you can use the limit-direct-substitution theorem to evaluate the limit. – coreyman317 Oct 11 '18 at 11:15
• Note also that $x^2 + xx_0 + x_0^2$ is continuous at $x = x_0$. – Theo Bendit Oct 11 '18 at 11:18
• @coreyman317 that helped me! Thank you! – Recca Oct 11 '18 at 11:26

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

• Still gives me the same answer – Recca Oct 11 '18 at 11:18
• Yeah I didn't know you could substitute the limits. I solved it just now. Thanks anyways! – Recca Oct 11 '18 at 11:30
• I wish you a nice day! – Dr. Sonnhard Graubner Oct 11 '18 at 11:32
• Thanks! And you too. – Recca Oct 11 '18 at 11:46

Polynomial division:

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

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

$$-------$$

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

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

$$-------$$

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

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

$$--------$$

$$\small{0}$$