The basic rules of Differentiation of functions in calculus are presented along with several examples .

1 - Derivative of a constant function.

\( \)\( \)\( \)\( \)
The derivative of \( f(x) = c \) where c is a constant is given by
\[ f '(x) = 0 \]
Example
Given \( f(x) = - 10 \) , hence\( f '(x) = 0 \)

2 - Derivative of a power function (power rule).

The derivative of \( f(x) = x^r \) where \( r \) is a constant real number is given by
\[ f '(x) = r \; x^{r-1} \]
Example
Given\( f(x) = x^{-2} \) , hence \( f '(x) = -2 x^{-2-1} = \dfrac{-2}{x^3} \)

3 - Derivative of a function multiplied by a constant.

The derivative of \( f(x) = c \; g(x) \), where \( c \) is a constant, is given by
\[ f '(x) = c \; g '(x) \]
Example
Given \( f(x) = 3 \; x^3 \) ,
let \( c = 3 \) and \( g(x) = x^3 \), hence \( f(x) = c \; g(x) \)
and \( f '(x) = c \; g '(x) = 3 (3 x^2) = 9 \; x^2 \)

4 - Derivative of the sum of functions (sum rule).

The derivative of \( f(x) = g(x) + h(x) \) is given by
\[ f '(x) = g '(x) + h '(x) \]
Example
Given \( f(x) = x^2 + 4 \)
let \( g(x) = x^2 \) and \( h(x) = 4 \) Hence \( f '(x) = g '(x) + h '(x) = 2 x + 0 = 2 x \)

5 - Derivative of the difference of functions.

The derivative of \( f(x) = g(x) - h(x) \) is given by
\[ f '(x) = g '(x) - h '(x) \]
Example
Given \( f(x) = x^3 - x^{-2} \)
let \( g(x) = x^3 \) and \( h(x) = x^{-2} \).
Hence
\( f '(x) = g '(x) - h '(x) = 3 x^2 - (-2 x^{-3}) = 3 x^2 + 2 x^{-3} \)

6 - Derivative of the product of two functions (product rule).

The derivative of \( f(x) = g(x) \cdot h(x) \) is given by
\[ f '(x) = g(x) \cdot h '(x) + h(x) \cdot g '(x) \]
Example
Given \( f(x) = (x^2 - 2x) (x - 2) \)
let \( g(x) = (x^2 - 2x) \) and \( h(x) = (x - 2) \).
Hence
\( f(x) = g(x) \cdot h(x) \)
and using the formula, we obtain
\( f '(x) = g(x) h '(x) + h(x) g '(x) = (x^2 - 2x) (1) + (x - 2) (2x - 2) \)
Expand and group
\( f '(x) = x^2 - 2x + 2 x^2 - 6x + 4 = 3 x^2 - 8 x + 4 \)

7 - Derivative of the quotient of two functions (quotient rule).

The derivative of \( f(x) = \dfrac{g(x)}{h(x)} \) is given by
\[ f '(x) = \dfrac{h(x) g '(x) - g(x) h '(x)}{ (h(x))^2} \]
Example
Given \( f(x) = \dfrac{x-2}{x+1} \)
let \( g(x) = x - 2 \), \( \; h(x) = x + 1 \), hence \( f(x) = \dfrac{g(x)}{h(x)} \), \( g '(x) = 1 \) and \( h '(x) = 1 \).
Use the formula given above
\( f '(x) = \dfrac{h(x) g '(x) - g(x) h '(x)}{ (h(x))^2} \)
Substitute \( h(x), g(x), h'(x) \) and \( g'(x) \) by their expressions
\( f '(x) = \dfrac { (x + 1)(1) - (x - 2)(1) } {(x + 1)^2} \)
Group and simplify
\( f '(x) = \dfrac{3}{(x + 1)^2} \)