Derivative at a point

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 $\displaystyle \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}- \ldots$