# The Derivative function

If we want to have information of the derivative in all the points simultaneously, one cannot calculate the derivative at each point, since there are infinite points in any given interval. It is necessary to resort then to the derivative function.

The derivative function assigns to every point $x$ the value of the derivative at this point. It is defined as follows: $$\displaystyle f'(x) = \lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

See the following example:

$\displaystyle f(x)=x^2$

$$f'(x)=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x}=\lim_{\Delta x \to 0}\frac{\Delta x^2+2x\Delta x}{\Delta x}=$$ $$=\lim_{\Delta x \to 0}\Delta x+2x=2x$$ Therefore, $f '(x) = 2x$.

Given $f(x)=x(3x-1)$ we can compute the derivative as follows: $$\displaystyle f'(x)=\lim_{\Delta x \to 0}\frac{(x+\Delta x)(3(x+\Delta x)-1)-x(3x-1)}{\Delta x}=$$ $$=\lim_{\Delta x \to 0}\frac{3(x+\Delta x)^2-(x+\Delta x)-3x^2+x}{\Delta x}=$$ $$=\lim_{\Delta x \to 0}\frac{3(x^2+2x\Delta x+\Delta x^2)-3x^2-\Delta x}{\Delta x}=$$ $$=\lim_{\Delta x \to 0}\frac{6x\Delta x+3\Delta x^2-\Delta x}{\Delta x}=6x-1$$