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Let the function :

$$ f(x) =x^{3}-3x $$ with its domain in the real numbers.

Determine with the help of $$ f'(x) \equiv \lim _{h\rightarrow 0}\dfrac{f\left( x+h\right) -f\left( x\right) }{h} $$

the derivative $f'$ of function $f$.

I tried to plug the values of $f$ inside of $f'$:

$$ f (x ) =\lim _{h \to 0}\frac{(x+h)^{3}-3(x+h) -(x^{3}-3x) }{h} $$

I then tried to factorize but it didn't yeld results.

I don't understand how I can expand $(x + h)^3$ properly.

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    $\begingroup$ Expand instead of factor. You will get some nice cancellation. $\endgroup$
    – Clayton
    Sep 18, 2021 at 22:18
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    $\begingroup$ Can you expand $(x+h)^3$? $\endgroup$
    – Joe
    Sep 18, 2021 at 22:19
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    $\begingroup$ $(x+h)^3=(x+h)^2(x+h)=(x+h)(x+h)(x+h)$ use which ever is easier for you to expand it, you also have binomial formula but I'd try to get comfortable with expanding first. $\endgroup$
    – kingW3
    Sep 18, 2021 at 22:54

3 Answers 3

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In this particular case, I think it is more interesting to consider the following equivalent definition: \begin{align*} f'(a) & = \lim_{x\to a}\frac{f(x) - f(a)}{x-a}\\\\ & = \lim_{x\to a}\frac{x^{3} - 3x - a^{3} + 3a}{x - a}\\\\ & = \lim_{x\to a}\frac{(x-a)(x^{2} + ax + a^{2}) - 3(x-a)}{x-a}\\\\ & = \lim_{x\to a}\left(x^{2} + ax + a^{2} - 3\right)\\\\ & = 3a^{2} - 3 \end{align*}

where we have made use of the identity:

\begin{align*} x^{3} - y^{3} = (x-y)(x^{2} + xy + y^{2}) \end{align*}

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  • $\begingroup$ Could you explain more in detail what the a is ? $\endgroup$
    – WindBreeze
    Sep 18, 2021 at 23:22
  • $\begingroup$ The number $a$ is some point of the domain of $f$ which is also an accumulation point. $\endgroup$ Sep 18, 2021 at 23:25
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Look that if you expand your numerator you get:

$$\lim _{h\to 0} \frac{(x+h)^3-3(x+h)-(x^3-3x)}{h}=\lim _{h\to 0}(3x^2+3xh+h^2-3)=3x^2-3$$

Remember that: $(x+h)^3=x^3+3x^2h+3xh^2+h^3$ reading the comment of the user.

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  • $\begingroup$ Now try it with the epsilon delta definition. $\endgroup$ Sep 18, 2021 at 23:04
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I give you a hint: expand everything, see what cancels, and then divide by $h$ and then take the limit. With that, you should get the required solution.

For your second question regarding expansion, I think it is best to remember the formula for $n\in\mathbb{N}$:

$$(x+h)^n=\displaystyle\sum_{i=0}^n {n\choose i}x^ih^{n-i}$$

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  • $\begingroup$ My problem is I have difficulty expanding everything because of the cube exponent. $\endgroup$
    – WindBreeze
    Sep 18, 2021 at 22:19
  • $\begingroup$ @WindBreeze Well, do you know the binomial theorem for expansion? $\endgroup$ Sep 18, 2021 at 22:20
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    $\begingroup$ @WindBreeze, do you know how to expand $(x+h)^2$? $\endgroup$
    – Joe
    Sep 18, 2021 at 22:20
  • $\begingroup$ @WindBreeze I have edited for you the binomial theorem for expansion, I think this might help you. $\endgroup$ Sep 18, 2021 at 22:49

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