In the definition of average variation we emphasize that we are considering any interval $$[a,a+\Delta x]$$. One can wonder what happens when we make this interval infinitely small. That is, being at any given point $$a$$, what hapens to the average change when the width of the interval $$\Delta x$$ becomes infinitely small?. The result is the definition of a derivative at point $$a$$.

$$$\displaystyle f'(a)=\lim_{\Delta x \to 0}\frac{f(a+\Delta x)-f(a)}{\Delta x}$$$

(It can be read: The derivative at the point $$a$$ is equal to the limit when $$\Delta x$$ tends to zero of the division $$\displaystyle \frac{f(a+\Delta x)-f(a)}{\Delta x}$$).

Following this, it is possible to look for the derivative of the function $$y=x^3$$ at the point $$a=3$$.

We only have to apply the definition: $$$\displaystyle y=f(x)=x^2 \Rightarrow f'(3)= \lim_{\Delta x \to 0}\frac{f(3+\Delta x)-f(3)}{\Delta x}=$$$

$$$\displaystyle=\lim_{\Delta x \to 0}\frac{(3+\Delta x)^2-3^2}{\Delta x}=\lim_{\Delta x \to 0}\frac{(9+6\Delta x+\Delta x^2)-9}{\Delta x}=$$$

$$$=\lim_{\Delta x \to 0}\frac{\Delta x^2+6\Delta x}{\Delta x}=\lim_{\Delta x \to 0}\Delta x+6=6$$$ When we take the limit as $$\Delta x$$ goes to $$0$$, the result is: $$f '(3) = 6$$.

Let's see some examples.

Find the derivative at $$x=0$$ of the following functions:

$$f(x)=x(3x-1)$$

$$$\displaystyle f'(0)=\lim_{\Delta x \to 0}\frac{(0+\Delta x)(3(0+\Delta x)-1)-0(3 \cdot 0-1)}{\Delta x}=\lim_{\Delta x \to 0}\frac{3\Delta x^2-\Delta x}{\Delta x}=$$$ $$$=\lim_{\Delta x \to 0}3\Delta x-1=-1$$$

$$f(x)=\sin x $$

$$$\displaystyle f '(x)=\lim_{\Delta x \to 0}\frac{f(0+\Delta x)-f(0)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\sin \Delta x-\sin0}{\Delta x}=\lim_{\Delta x \to 0}\frac{\sin\Delta x }{\Delta x}=1$$$

In the last instance it must be known that $$\displaystyle \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}- \ldots$$