# How do we calculate the derivative of $f(x) = x^{3} - 3x$?

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.

• Expand instead of factor. You will get some nice cancellation. Sep 18, 2021 at 22:18
• Can you expand $(x+h)^3$?
– Joe
Sep 18, 2021 at 22:19
• $(x+h)^3=(x+h)^2(x+h)=(x+h)(x+h)(x+h)$ 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. Sep 18, 2021 at 22:54

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*}

• Could you explain more in detail what the a is ? Sep 18, 2021 at 23:22
• The number $a$ is some point of the domain of $f$ which is also an accumulation point. Sep 18, 2021 at 23:25

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.

• Now try it with the epsilon delta definition. Sep 18, 2021 at 23:04

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 $$n\in\mathbb{N}$$:

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

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