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!

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    $\begingroup$ 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. $\endgroup$ – coreyman317 Oct 11 '18 at 11:15
  • $\begingroup$ Note also that $x^2 + xx_0 + x_0^2$ is continuous at $x = x_0$. $\endgroup$ – Theo Bendit Oct 11 '18 at 11:18
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    $\begingroup$ @coreyman317 that helped me! Thank you! $\endgroup$ – 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 $$3x_0^2$$

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

Polynomial division:

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

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


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

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







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