When we start with calculus, we studied about the functions, its types, domain range etc.  A function can be continuous or discontinuous at a particular point or set of points. A continuous function can be either differentiable or not. In this section we are going to learn about the difference quotient, difference quotient of a function and the limit of difference quotient. This will help us to find the derivative of difference functions like, trigonometric, exponential,  logarithmic, hyperbolic functions.

Differentiation help us to find the velocity and acceleration of a moving particle when the distance is given as a function of a variable. Any other function which depend on the rate of change can also be calculated if we are comfortable in finding the derivative of a function. This section will help us to understand the basics of differentiability.

## Finding the Difference Quotient

Let us consider any two points ( x1, y1 ) and ( x2 , y2 ) on the two dimensional plane.
The following diagram shows the difference quotient between the two points, which is the slope of the line joining the two points.
Difference Quotient = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

In the above figure, In right triangle PQR, $\angle$ PQR = $\theta$

Tan $\theta$ = $\frac{PR}{QR}$

= $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

( i. e ) The slope of the line joining the points P and Q is $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

### Solved Example

Question: Find the difference quotient of the function y = 4 x2 - 4 between the points x = 1 and x = 3
Solution:

We have y = 4 x2 - 4
Let f ( x ) =  4 x2 - 4
at x = 1, f ( 1 ) = 4 - 4 = 0
at x = 3, f ( 3 ) = 4 ( 3 )2 - 4
= 36 - 4
= 32
Therefore we have y1 = 0 and y2 = 32
The Difference Quotient = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{32-0}{3-1}$

= $\frac{32}{2}$

Difference Quotient = 16

## Difference Quotient of a Function

Let y = f ( x ) be a given function and ( x1, y1 ) be a point on the graph of the function.

Let h be a small positive number less than 1, such that x is given a small increment x + h, then the value of y also increases as described below.

Let ( x1, y1 ) be a point on the given function and ( x2, y2 ) be the new point obtained when x is given a small increment.

As both the points satisfy the function y = f ( x ), we get,
y1 = f ( x1 )
and y2 = f ( x2 )
where x2 = x1 + h, where h is the increment given to the point x1 .

Difference quotient = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{f\left ( x_{2} \right )-f\left ( x_{1} \right )}{x_{2}-x_{1}}$

= $\frac{f\left ( x_{1}+h \right )-f\left ( x_{1} \right )}{x_{1}+h-x_{1}}$  [ Substituting x2 = x1 + h ]

= $\frac{f\left ( x_{1}+h \right )-f\left ( x_{1} \right )}{h}$

Therefore the difference quotient of the function y = f ( x) is, $\frac{f\left ( x_{1}+h \right )-f\left ( x_{1} \right )}{h}$

The difference in quotient at a point x = a is given by,  $\frac{f(a+h)-f(a)}{h}$

For any arbitrary point ( x, y ) the Difference Quotient of the function y = f ( x) is given by, $\frac{f(x+h)-f(x)}{h}$

### Solved Example

Question: Find the difference Quotient of the function f ( x ) = 5 x2 - 6 x + 9, at x = -2
Solution:

We have f ( x ) = 5 x2 - 6 x + 9
The Difference Quotient of a function at x = a is $\frac{f(a+h)-f(a)}{h}$

Therefore, f ( - 2 + h) = 5 ( -2 + h)2 - 6 ( -2 + h) + 9

= 5 ( 4 - 4h + h2 ) + 12 - 6h + 9

= 20 - 20 h + 5 h2 + 21 - 6 h

= 5 h2 - 26 h + 41

f ( -2 ) = 5 ( -2 )2 - 6 ( -2 ) + 9

= 20 + 12 + 9

= 41
f ( a + h ) - f ( a ) = f ( -2 + h ) - f ( -2 )

= 5 h2 - 26 h + 41 - 41

= 5 h2 - 26 h

= h ( 5 h - 26 )

$\frac{f(a+h)-f(a)}{h}$ = $\frac{h(5h-26)}{h}$

$\frac{f(a+h)-f(a)}{h}$   = 5h - 26

## Symmetric Difference Quotient

Let the function be y = f ( x ) and ( x, y ) be any arbitrary point P on the graph of the given function y = f ( x ).

Let h be a small positive value < 1.

If P approaches Q from right ( towards left ), the difference quotient from P to Q is given by, $\frac{f(x)-f(x-h)}{h}$

If P approaches R towards right, the difference in quotient from P to R is given by, $\frac{f(x+h)-f(x)}{h}$

The symmetric difference quotient is the difference quotient from Q to R given by
$\frac{f(x+h)-f(x-h)}{x+h-(x-h)}$

= $\frac{f(x+h)-f(x-h)}{x+h-x+h}$

= $\frac{f(x+h)-f(x-h)}{h+h}$

= $\frac{f(x+h)-f(x-h)}{2h}$, which is the average of difference quotient from P to Q and R to P

The symmetric Difference Quotient is the difference quotient around the point x, in the interval ( x - h, x + h ) and is given by $\frac{f(x+h)-f(x-h)}{2h}$

### Solved Example

Question: Find the difference quotient around x for the function f ( x ) = $\frac{2x+3}{3x-5}$
Solution:

We have f ( x ) = $\frac{2x+3}{3x-5}$

According to the above discussion, Symmetric difference quotient  = $\frac{f(x+h)-f(x-h)}{2h}$, where 0 < h < 1
f(x + h) = $\frac{2(x+h)+3}{3(x+h)-5}$

= $\frac{2x+2h+3}{3x+3h-5}$

f(x - h) = $\frac{2(x-h)+3}{3(x-h)-5}$

= $\frac{2x-2h+3}{3x-3h-5}$

f(x + h) - f ( x - h ) = $\frac{2x+2h+3}{3x+3h-5}$ - $\frac{2x-2h+3}{3x-3h-5}$

= $\frac{(2x+2h+3)(3x-3h-5)-(2x-2h+3)(3x+3h-5)}{(3x+3h-5)(3x-3h-5)}$

$\frac{f(x+h)-f(x-h)}{2h}$  =

$\frac{(2x+2h+3)(3x-3h-5)-(2x-2h+3)(3x+3h-5)}{2h(3x+3h-5)(3x-3h-5)}$

## Limit of the Difference Quotient

For the function y = f ( x ), we have the difference Quotient $\frac{f(x+h)-f(x)}{h}$, where 0 < h < 1.

The difference quotient is the slope of the line joining the two points, which is the chord of the circle.
As h -> 0, the two points come closure and become the tangent of the curve at the point ( x, y )
Therefore, the slope of the chord will be the same as the slope of the tangent at the point ( x, y ) which is given as, $\lim_{h->0}\frac{f(x+h)-f(x)}{h}$ which is called as the derivative of the function f ( x ) denoted as f ' ( x ).
At a given point x = a, the derivative of the function is written as $\lim_{h->0}\frac{f(a+h)-f(a)}{h}$ which is denoted as f ' ( a ).

### Solved Example

Question: Find the derivative of the function y = 2 x2 + 3x - 9, at x = 3, using the limit of difference quotient.
Solution:

We have y = f ( x ) = 2 x2 + 3x - 9,
x = 3 is the given point and let h be a small increment given to 3, such that 0 < h < 1.
According to the above difference quotient formula at x = a, we have,

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

= $\lim_{h->0}\frac{f(3+h)-f(3)}{h}$ - - - - - -  - - - - - ( 1 )

f ( x ) = 2 x2 + 3 x - 9

f ( 3 ) = 2 ( 3 )2 + 3 ( 3 ) - 9 = 18 + 9 - 9 = 18

f ( 3 + h ) = 2 ( 3 + h )2 + 3 ( 3 + h ) - 9

= 2 ( 9 + 6 h + h2 ) + 9 + 3 h - 9

= 2 h2 + 15 h + 18

Substituting for f ( 3 ) and f ( 3 + h ) in  ( 1 ), we get,

f ' ( 3 ) = $\lim_{h->0}$  $\frac{f(3+h)-f(3)}{h}$

= $\lim_{h->0}$  $\frac{2h^{2}+15h+18-18}{h}$

= $\lim_{h->0}$  $\frac{2h^{2}+15h}{h}$

= $\lim_{h->0}$  $\frac{h(2h+15)}{h}$

= $\lim_{h->0}$  (2h + 15 )

= 0 + 15

f ' ( 3 ) = 15

## Difference Quotient Examples

### Solved Examples

Question 1: Find the Difference Quotient of the function f ( x ) = sin x.
Solution:

We have f ( x ) = sin x
The Difference Quotient = $\frac{f(x+h)-f(x)}{h}$,  where 0 < h < 1

f ( x + h ) - f ( x ) = sin ( x + h ) - sin x

=  2 cos ( $\frac{x+h+x}{2}$ ) sin ( $\frac{x+h-x}{2}$ )

= 2 cos (x + $\frac{h}{2}$) . sin ($\frac{h}{2}$)

$\frac{f(x+h)-f(x)}{h}$  = $\frac{cos(x+\frac{h}{2}).sin(\frac{h}{2})}{h}$

Question 2: Find the limit of Difference Quotient of the function f ( x ) = 3 x2 - 5 x + 8, at x = 2
Solution:

We have f ( x) = 3 x2 - 5 x + 8
Limit of Difference Quotient = $\lim_{h->0}$  $\frac{f(a+h)-f(a)}{h}$, at x = a

= $\lim_{h->0}$  $\frac{f(2+h)-f(2)}{h}$, at x = 2

f ( x )     = 3 x2 - 5 x + 8

f ( 2 + h ) = 3 ( 2 + h )2 - 5 ( 2 + h ) + 8

= 3 ( 4 + 4 h + h2 ) - 10 - 5h + 8

= 12 + 12 h + 3 h2 - 10 - 5 h + 8

= 3 h2 + 7 h + 10

f ( 2 ) = 3 ( 2 )2 - 5 ( 2 ) + 8 = 12 - 10 + 8 = 10

f ( 2 + h ) - f ( 2 ) = 3 h2 + 7 h + 10  - 10 = 3 h2 + 7 h = h ( 3h + 7 )

$\frac{f(2+h)-f(2)}{h}$ = $\frac{h(3h+7)}{h}$

= 3 h + 7

Therefore, $\lim_{h->0}$  $\frac{f(2+h)-f(2)}{h}$

= $\lim_{h->0}$  ( 3h + 7 )

= 3 ( 0 ) + 7

$\lim_{h->0}$  $\frac{f(2+h)-f(2)}{h}$ = 7