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.

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.
