A derivative is the instantaneous rate of change of a function. For example while finding rates of change, the instantaneous rate of change of a function f(x) when x = a is equal to the derivative of f at a.

If f(x) = y and for a small change in x say $\partial$x , the function changes by $\partial$y. If the
$\lim_{\partial x \rightarrow 0}$ $\frac{\partial y}{\partial x}$ exists then it is called as the derivative of the function.
Derivatives have their own notation. The derivative of a function f at a number a is denoted by f'(a) and is given by:

$f'(a) = $$\lim_{h \rightarrow 0}$$\frac{f(a+h) - f(a)}{h}$ = $\lim_{x \rightarrow a}$$\frac{f(x) - f(a)}{x - a}$

The most common notation to denote derivative is $f'(x)$. Some other notations are
$f'(x) = y'$ = $\frac{dy}{dx}$ = $\frac{d}{dx}$ $f(x)$

The derivative of a function f(x) with respect to x is defined as

$f'(x)$ = $\lim_{h \rightarrow 0}$$\frac{f(x + h) - f(x)}{h}$

Solved Example

Question: Solve the derivative of the given function by using derivative definition.
$f(x)$ = $x^2$ + 5x
Solution:
 
$ƒ(x)$ = $x^2$ + 5x
ƒ(x+h)  = $(x+h)^2$ + 5(x+h)
              = $x^2 + 2hx + h^2 + 5x + 5h$
              
Apply the definition of derivative

    $ƒ'(x)$ = $\lim_{h \rightarrow 0}$ $\frac{f(x + h) - f(x)}{h}$
    
         So, $ƒ'(x)$ = $\lim_{h \rightarrow 0}$$\frac{x^2 + 2hx + h^2 + 5x + 5h - (x^2 + 5x)}{h}$
        
$\lim_{h \rightarrow 0}$$\frac{2hx + h^2 + 5h}{h}$

Take h common from numerator.

= $\lim_{h \rightarrow 0}$$\frac{h(2x + h + 5)}{h}$

Cancel out h and simplify

$ƒ'(x)$ = $\lim_{h \rightarrow 0}$ 2x + h + 5

         Plug in the limit h=0,
        
$ƒ‘(x)$ = 2x + 0 + 5 = 2x + 5.