what are derivatives?

A derivative of a function at any given point is its instantaneous rate of change at that point. You can also think of the derivative of a function as the graph of the slopes of that function.

notations

1. prime notation: $f'(x)$ (or $y'$) is the derivative of $f(x)$ (or $y$).

2. delta notation: $\frac {d} {dx} f(x)$ (or $\frac {dy} {dx}$) is the derivative of $f(x)$ (or $y$).

visualization

The following video demonstrates how the derivative of a function is determined, by first calculating the slope at a given point and then plotting it onto the graph:

Note that the straight line on the $f(x)$ curve is the tangent of that curve.

Also, notice that when the value of the derivative gets closer to zero, the tangent becomes more flat. This makes sense, since the deriviative of a function is the value of the slope at a particular point.

Determining the derivative of a function

As of now, we can only find the derivative of a function at a given point, but what if we want the actual function that represents the derivative of a function?

We can do this by using the formula for the instantaneous rate of change at a point, but instead of treating $a$ as a point, we use $x$ instead:

Derivative by first principle

$f'(x) = \frac {dy} {dx} = \ ^{\lim}_{h \to 0} \frac {f(x + h) - f(x)} {h}$

example

Determine the derivative of the following function: $f(x) = x^2 + x$

answer

Using the first principle method, we can calculate the derivative:

$f'(x) = \ ^{\lim}_{h \to 0} \frac {f(x + h) - f(x)} {h}$

$\quad \quad \ \ = \ ^{\lim}_{h \to 0} \frac { [(x + h)^2 + (x + h)] - [x^2 + x]} {h}$

$\quad \quad \ \ = \ ^{\lim}_{h \to 0} \frac { x^2 + 2hx + h^2 + x + h - x^2 - x} {h}$

$\quad \quad \ \ = \ ^{\lim}_{h \to 0} \frac { \cancel{x^2} + 2hx + h^2 \cancel{ + x} + h \cancel{ - x^2} \cancel{ - x}} {h}$

$\quad \quad \ \ = \ ^{\lim}_{h \to 0} \frac { 2hx + h^2 + h } {h}$

$\quad \quad \ \ = \ ^{\lim}_{h \to 0} \frac { h (2x + h + 1) } {h}$

$\quad \quad \ \ = \ ^{\lim}_{h \to 0} \frac { \cancel{h} (2x + h + 1) } {\cancel{h}}$

$\quad \quad \ \ = \ ^{\lim}_{h \to 0} (2x + h + 1)$

$\quad \quad \ \ = 2x + (0) + 1$

$\quad \quad \ \ = 2x + 1$

$\therefore f'(x) = 2x + 1$