rules of finding derivatives of different types of functions in mathematics

 To find out derivatives of different types of functions, you need to follow some rules. They are sum/difference rule, product rule, chain rule and quotient rule. 

See details of these rules from HERE.

See the detailed concepts of derivatives, HERE.

Here we demonstrate these rules with the help of examples.

1. Sum/difference rule:

This rule is used when the function is in the form of sum or difference. In other words, sum rule is useful to find the derivatives when the function is in polynomial form, that is,

`f(x) = ax \pm bx^2 \pm cx^3 + .........` 

Let's  consider `y = f(x) = u(x) \pm v(x) \pm w(x)`, then derivative of f(x) is,

`\frac{d f(x)}{dx} = \frac{d u(x)}{dx} \pm \frac{d v(x)}{dx} \pm \frac{d w(x)}{dx}`

Examples:

(i) `f(x) = 3x^3 - 2x + 1`

We need to find derivative of f(x).

`\frac{d}{dx}f(x) = \frac{d}{dx}(3x^3-2x+1)`

               = `\frac{d}{dx}3x^3 - \frac{d}{dx}2x + \frac{d}{dx}1`

              = `3 \times 3x^2 - 2 \times 1` + 0`

             = `9x^2 -2`

`\therefore` `\frac{d}{dx} (3x^3-2x+1) = 9x^2 -2` // 

2. Product rule

This rule is used to determine the derivative when the function is in product form. 

That is, when `f(x) = u(x) \cdot v(x) \cdot w(x)`

Then derivative of f(x) w.r.t x is,

`\frac{d}{dx} f(x)= v(x) w(x) \frac{d}{dx} u(x) + u(x) w(x) \frac{d}{dx} v(x) + u(x) v(x) \frac{d}{dx} w(x)`.

Example:

(i) `f(x) = 3x^2(2x-1)`

Let's find out derivative of f(x) w.r.t x,

`\frac{d}{dx}f(x) = \frac{d}{dx} 3x^2(2x-1)`

Let's `u(x) = 3x^2`, v(x) = 2x-1, then

`\frac{d}{dx}3x^2.(2x-1) = (2x-1)\frac{d}{dx}3x^2 + 3x^2\frac{d}{dx}(2x-1)`

or, `\frac{d}{dx}f(x) = (2x-1) 3 \times 2 x + 3x^2 \times (2 \times - 0)`

or, `\frac{d}{dx}f(x) = 6x(2x-1) + 6x^2`

or, `\frac{d}{dx}f(x) = 12x^2 - 6x + 6x^2`

or, `\frac{d}{dx}f(x) = 18x^2 - 6x = 6x(3x-1)`

`\therefore` `\frac{d}{dx}[3x^2.(2x-1)]= 6x(3x-1)` //

3. Power rule

This rule is used to find derivative of those functions which contain higher or fractional power terms. 

If `f(x) = u^n(x)` ,then

`\frac{d}{dx} u^n(x) = n u^{n-1}(x) \frac{u(x)}{dx}`

Example:

(i) `f(x) = (x^{\frac{1}{2}} + x^-\frac{1}{2})^2`

Let's find the derivative of this function by using the power rule.

`\frac{d}{dx}f(x) = \frac{d}{dx} (x^\frac{1}{2}+x^-\frac{1}{2})^2`

or, `\frac{d}{dx}f(x) = \frac{d}{d(x^\frac{1}{2}+x^-\frac{1}{2})}(x^\frac{1}{2}+x^-\frac{1}{2})^2 \cdot \frac{d}{dx}(x^\frac{1}{2}+x^-\frac{1}{2})`

or, `\frac{d}{dx}f(x) = 2(x^\frac{1}{2}+x^-\frac{1}{2})^(2-1) \cdot (\frac{d}{dx} x^\frac{1}{2} + \frac{d}{dx}x^-\frac{1}{2})` 

or, `\frac{d}{dx}f(x) = 2(x^\frac{1}{2}+x^-\frac{1}{2}) \cdot (\frac{1}{2}x^(\frac{1}{2}-1)-\frac{1}{2}x^(-\frac{1}{2}-1))`

or, `\frac{d}{dx}f(x) = 2(x^\frac{1}{2}+x^-\frac{1}{2}) \cdot (\frac{1}{2} x^-\frac{1}{2}-\frac{1}{2}x^-\frac{3}{2})`

or, `\frac{d}{dx}f(x) = (x^\frac{1}{2}+x^-\frac{1}{2}) \cdot (x^-\frac{1}{2}-x^-\frac{3}{2})`

or, `\frac{s}{dx}f(x) = x^\frac{1}{2} \cdot x^-\frac{1}{2} - x^\frac{1}{2} \cdot x^-\frac{3}{2}+x^-\frac{1}{2} \cdot x^-\frac{1}{2}-x^-\frac{1}{2} \cdot x^-\frac{3}{2}`

or, `\frac{d}{dx}f(x) = 1 - x^-1 + x^-1 - x^-2`

or, `\frac{d}{dx}f(x) = 1 - x^-2`

`\therefore` `\frac{d}{dx} (x^\frac{1}{2}+x^-\frac{1}{2})^2 = 1 - \frac{1}{x^2}` //

4. Chain rule

This rule is used to determine the derivative of function which is in explicit form. This rule is somehow similar to the power rule.

If f(u) = u(x) and u(x) = x(t) then,

`\frac{d}{dx}f = \frac{d}{d x}u(x) \cdot \frac{d}{dt}x(t)` 

Example:

(i) `f(u) = 2u^2+3` and `u =  3x^2 - 1` 

`\frac{d}{dx}f = \frac{d}{du) f(u) \cdot \frac{d}{dx}u(x)`

or, `\frac{d}{dx}y = \frac{d}{du}(2u^2+3) \cdot \frac{d}{dx}(3x^2-1)`

or, `\frac{d}{dx}y = (4u) \cdot (6x)`

or, `\frac{d}{dx}y = 4(3x^2-1)(6x)`        [`\therefore` Use of u(x) ]

or, `\frac{d}{dx}y = 24x(3x^2-1)`

`\therefore` `\frac{d}{dx} y = 24x(3x^2-1)` //

5. Quotient rule

This rule is used to determine derivatives when the function is in the quotient form or in fractional form.

If `f(x) = \frac{u(x)}{v(x)}` then

`\frac{d}{dx}f(x) = \frac{v(x) \frac{d}{dx}u(x) - u(x) \frac{d}{dx}v(x)}{v^2(x)}`.

Example:

(i) `f(x) = \frac{x^2}{1-x^2}`

Let's `u (x) = x^2` and `v(x) = 1-x^2` then 

`\frac{d}{dx}f(x) = \frac{d}{dx}(\frac{u(x)}{v(x)})`

            = `\frac{d}{dx} (\frac{x^2}{1-x^2})`

           = `\frac{(1-x^2) \frac{d}{dx}x^2 - x^2 \frac{d}{dx}(1-x^2)} {(1-x^2)^2}`

         = `\frac{(1-x^2) 2x - x^2 (-4x)}{(1-x^2)^2}`

       = `\frac{2x(1-x^2) + 4x^3}{(1-x^2)^2}`

      = `\frac{2x - 4x^3 + 4x^3}{(1-x^2)^2}`

`\therefore` `\frac{d}{dx}(\frac{x^2}{1-x^2}) = \frac{2x}{(1-x^2)^2}` //

If you are interested in trigonometric functions, then for learning to find out derivatives of all trigonometric functions, Click HERE.