Differentiation is one of the basic branches of Calculus. It describes the real world rates of change and helps us describe the physical universe and natural phenomena in terms of Mathematics.

Rate of change involves 2 or more variables:

- Dependent variables
- Independent variables

**In differential Calculus**, we mainly deal with the rate of change of a dependent variable with respect to one or more independent variables.
We first explain the terms dependent and independent variables.

We usually write `y = f(x)` where `f(x)` is the value of `f ` at ` x ԑ D ` (the Domain of the function f). Let us consider the functional relation
`y = f(x) = x^2 + 1`

For different values of `x ԑ D, f(x)` or the expression `x^2 + 1` assumes different values. For example; if `x = 1, 1.5, 2` etc., then

`f(1) =(1)^2 + 1 = 2,`

`f(1.5) `=`(1.5)^2 + 1` = `2.25 + 1 `= `3.25,`

`f(2) =(2)^2 + 1 = 5`

`f(1.5) `=`(1.5)^2 + 1` = `2.25 + 1 `= `3.25,`

`f(2) =(2)^2 + 1 = 5`

We see that for the change `1.5 – 1 = 0.5` in the value of x, the corresponding change in the value of `y ` or ` f(x)` is given by
`f(1.5) – f(1)` =` 3.25 – 2 `= `1.25`

It is obvious that the change in the value of the expression `x^2 + 1 (or f(x)) `depends upon the change in the value of the variable `x`. As `x`
behaves independently, so we call it the **independent** variable. But the behavior of `y or f(x)` **depends** on the variable `x`, so we
call it the dependent variable.

The change in the value of `x` (positive or negative) is called the increment of `x` and is denoted by the symbol `δx` (read as delta x). The
corresponding change in the value of dependent variable `y or f(x)` for the change `δx` in the value of `x` is denoted by ` δyor δf = f(x + δx) – f(x)`.
Usually the small changes in the value of the variables are taken as increments of variables.

Suppose a particle (or an object) is moving in a straight line and its positions from (from some fixed point) after times t and t1 are given by` s(t)` and `s(t1)`, respectively. Then the distance traveled in the time interval `t1 – t` where `t1 > t ` is ` s(t1) – s(t)`.

The difference quotient`(s(t1) – s(t))/(t1 – t)`represents the average rate of change of distance over the time interval `t1 – t`.

If `t1 – t` is not small, then the average rate of change does not represent an accurate rate of change near `t`. If however `t1` approaches `t`, then

`Limt1 → t(s(t1) – s(t))/(t1 – t)`

Provided this limit exists, is called the **instantaneous** rate of change of s with respect to t at t and is written as s'(t).

If for some `x1 = x +δx ` i.e.,` x1 – x = δ` , then the limit

`Limδx→ 0(f(x+ δx) – f(x))/δx`

provided it exists is defined to be the **derivative of f. **A common notation used for derivative of a function `f(x)` is `f'(x)` (read as ` f-`
prime of ` x ` ).

Differentiation is the process of finding derivative. The two things are same, so do not be confused between the two.

`dy/dx` is a widely used notation for the derivative of a dependant variable y w.r.t. an independent variable

X. Also this derivative is represented by y'.We will be using all these notations in the future throughout this course content.

Depending on different types of functions, we can have various methods to find their derivatives. The most basic approach used for finding derivatives of
functions is as shown above, and is called **differentiation by definition.**

We present an example of differentiation that makes use of this method.

From the basic definition of derivative, we have

`f(x)=Limδx→ 0(f(x+ δx) – f(x))/δx`

`f(x)= x^2 ; f(x+ δx) = (x + δx )^2`

`f(x)=Limδx→ 0( ( x + δx )^2 – x^2)/δx`

=`Limδx→ 0(x^2 + ( δx )^2+ 2(x)(δx) – x^2)/δx`

=`Limδx→ 0( ( δx )^2+ 2(x)(δx))/δx= Limδx→ 0( 2x +δx)`

`f(x)= x^2 ; f(x+ δx) = (x + δx )^2`

`f(x)=Limδx→ 0( ( x + δx )^2 – x^2)/δx`

=`Limδx→ 0(x^2 + ( δx )^2+ 2(x)(δx) – x^2)/δx`

=`Limδx→ 0( ( δx )^2+ 2(x)(δx))/δx= Limδx→ 0( 2x +δx)`

Putting the limit `δx → 0 `into the expression

*`f(x) = 2x` *

that is the required derivative of the given function.

This is the most basic approach for finding the derivative of a function.

Since the definition of derivative itself is employed in this method, hence the name differentiation by definition.

Next we will learn some easier methods / theorems that are employed effectively to find derivatives of different types of functions.

Differentiation of `x^n` where n is an integer

`(d [x^n])/dx= n. x^(n-1)` where ` n ∈Z`

The above identity is called the

Power Rule,and it can be proved very easily using the ab-initio method of differentiation.

(Using the Power Rule)

` (d [x^n])/dx= n. x^(n – 1)`

` =>(d [x^2])/dx=2. x^(2 – 1 ) = 2x`

` =>(d [x^2])/dx=2. x^(2 – 1 ) = 2x`

that is the same result as obtained previously by ab-initio method. This shows that the Power Rule is consistent with the definition of derivative.

(According to the Power rule we studied above)

`(d [x^n])/dx= n. x^(n – 1)`

We have `f(x)= x^( -4 )`

`i.e., ` we have ` n = –4` ` =>f^' (x)=(-4) x^( -4-1)`

` =>f^' (x)=-4x^( -5)`

We have `f(x)= x^( -4 )`

`i.e., ` we have ` n = –4` ` =>f^' (x)=(-4) x^( -4-1)`

` =>f^' (x)=-4x^( -5)`

that is the required derivative of the given function.

`8/5x ^(–1/5)`

According to the theorem

`(d/dx (kf(x)) = k d/dx f(x) = kf^' (x))`

`f'(x)=d/dx(2x^(4/5)) =2d/dx( x^(4/5)) = 2. 4/5x^(4/5 – 1) = 8/5x ^(–1/5)`

`f'(x)=d/dx(2x^(4/5)) =2d/dx( x^(4/5)) = 2. 4/5x^(4/5 – 1) = 8/5x ^(–1/5)`

Please remember that the derivative of a constant function is zero. Mathematically it can be written as:** `d/dx(k) = 0`,where `k ∈Z.`**

Derivative of y = `kf(x)w.r.t. x`

`d/dx (kf(x)) = k d/dx f(x) = kf^' (x) ` where ` k ∈Z.`

`d/dx (kf(x)) = k d/dx f(x) = kf^' (x) ` where ` k ∈Z.`

Mathematically, we can write this as follows:The derivative of the Sum / Difference of two functions is given by the Sum / Difference of their derivatives.

` d/dx [u(x) + v(x)] `= `d/dx [u(x)] + d/dx [v(x)] `

Same is true for the difference of two functions:

` d/dx [u(x) - v(x)]` = `d/dx [u(x)] -d/dx [v(x)] `

We can extend this rule to find the derivative of the Sum and Difference of more than two functions.

We next present a few examples of these two simple theorems.

We first multiply the two terms in order to express the function ` f(x) ` in terms of the sum of sub-functions as shown below.

` f(x)= x^2 + 11x + 30 `

` f'(x)= d/dx[ x^2] + d/dx[11x ] + d/dx[30] `

` f'(x)= 2x + 11 `

` f'(x)= d/dx[ x^2] + d/dx[11x ] + d/dx[30] `

` f'(x)= 2x + 11 `

If two functions ` u(x) ` and ` v(x) ` can be differentiated at X, then their product ` u(x)v(x) ` can also be differentiated at X. And we can use the following rule to determine such a derivative.

` d/dx [u(x) . v(x)] ` =` d/dx [u(x)] .v(x) + d/dx [v(x)] . u(x) `

Note: The rule can be shown to be true by using the ab-initio method.

We are given that

` f(x)= (x+5)(x+6) `

Using the product of functions formula:

` d/dx [(x+5) . (x+6)] =d/dx [(x+5)] . (x+6) + d/dx [(x+6)] . (x+5) `

` f'(x)= (1)(x+6) + (1)(x+5) `

` f'(x)= x+6 + x+5 = 2x + 11 `

` f'(x)= (1)(x+6) + (1)(x+5) `

` f'(x)= x+6 + x+5 = 2x + 11 `

**Note: **

Results obtained in above two examples are same.

If two functions ` u(x) ` and ` v(x) ` can be differentiated at X, then their quotient ` (u(x))/(v(x)) ` can also be differentiated at X (provided ` v(x) ≠ 0 ` at ` x ` ). We can use the following rule to determine such a derivative.

` d/dx [(u(x))/(v(x))] ` = `(d/dx [u(x) ] .v(x)- d/dx [v(x) ].u(x))/([v(x)]2) `

We make use of the **Quotient Rule** to find the derivative of this function.

` d/dx [(u(x))/(v(x))] = (d/dx [u(x) ] .v(x)- d/dx [v(x) ].u(x))/([v(x)]2) `

Let ` u(x) = 2x^(3 )- 3x^(2 )+ 2 `

` v(x) = x^(2 )+ 1 `

` =>d/dx[( 2x^(3 )- 3x^(2 )+ 2)/( x^(2 )+ 1)]=(d/dx [ 2x^(3 )- 3x^(2 )+ 2] . (x^(2 )+ 1)- d/dx [( x^(2 )+ 1 )] ( 2x^(3 )- 3x^(2 )+ 2 ))/([( x^(2 )+ 1 )]2) `

` =([(3)(2x^2) – 3(2x)] . (x^(2 )+ 1) - [(2x)] ( 2x^(3 )- 3x^(2 )+ 2 ))/([( x^(2 )+ 1 )]2) `

` =( (6x^(2 )– 6x)(x^(2 )+ 1) – (2x)( 2x^(3 )– 3x^(2 )+ 2 ))/([( x^(2 )+ 1 )]2) `

` =((6x^4– 6x^3+6x^2-6x) – (4x^(4 )– 6x^(3 )+ 4x))/([( x^(2 )+ 1 )]2) `

` =>d/dx[( 2x^(3 )- 3x^(2 )+ 2)/( x^(2 )+ 1)]=( 6x^4- 6x^3+ 6x^2-6x – 4x^(4 )+ 6x^(3 )- 4x)/([( x^(2 )+ 1 )]2) `

` =>d/dx[( 2x^(3 )- 3x^(2 )+ 2)/( x^(2 )+ 1)]=( 2x^4+ 6x^2- 10x )/([( x^(2 )+ 1 )]2) `

Let ` u(x) = 2x^(3 )- 3x^(2 )+ 2 `

` v(x) = x^(2 )+ 1 `

` =>d/dx[( 2x^(3 )- 3x^(2 )+ 2)/( x^(2 )+ 1)]=(d/dx [ 2x^(3 )- 3x^(2 )+ 2] . (x^(2 )+ 1)- d/dx [( x^(2 )+ 1 )] ( 2x^(3 )- 3x^(2 )+ 2 ))/([( x^(2 )+ 1 )]2) `

` =([(3)(2x^2) – 3(2x)] . (x^(2 )+ 1) - [(2x)] ( 2x^(3 )- 3x^(2 )+ 2 ))/([( x^(2 )+ 1 )]2) `

` =( (6x^(2 )– 6x)(x^(2 )+ 1) – (2x)( 2x^(3 )– 3x^(2 )+ 2 ))/([( x^(2 )+ 1 )]2) `

` =((6x^4– 6x^3+6x^2-6x) – (4x^(4 )– 6x^(3 )+ 4x))/([( x^(2 )+ 1 )]2) `

` =>d/dx[( 2x^(3 )- 3x^(2 )+ 2)/( x^(2 )+ 1)]=( 6x^4- 6x^3+ 6x^2-6x – 4x^(4 )+ 6x^(3 )- 4x)/([( x^(2 )+ 1 )]2) `

` =>d/dx[( 2x^(3 )- 3x^(2 )+ 2)/( x^(2 )+ 1)]=( 2x^4+ 6x^2- 10x )/([( x^(2 )+ 1 )]2) `

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