# Rules of Differentiation of Functions in Calculus

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
$$f '(x) = \dfrac{3}{(x + 1)^2}$$