Derivative at a point

In the definition of average variation we emphasize that we are considering any interval $[a, a + Δ x]$ . 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 $Δ x$ becomes infinitely small?. The result is the definition of a derivative at point $a$ .

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

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

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: $y = f (x) = x^{2} \Rightarrow f^{'} (3) = lim_{Δ x \to 0} \frac{f (3 + Δ x) - f (3)}{Δ x} =$

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

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

Let's see some examples.

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

Example

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

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

Example

$f (x) = \sin x$

$f^{'} (x) = lim_{Δ x \to 0} \frac{f (0 + Δ x) - f (0)}{Δ x} = lim_{Δ x \to 0} \frac{\sin Δ x - \sin 0}{Δ x} = lim_{Δ x \to 0} \frac{\sin Δ x}{Δ x} = 1$

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