The difference quotient

Differentiation: The derivative

The difference quotient

Change plays a major role in the study of mathematics and in particular the study of functions. In the following example, we look at the average change at a chosen interval.

We calculate the average change on the interval #[2,4]# for the function #f(x)=2x-2#.

The horizontal change #\blue{\Delta x}# is: \[\blue{\Delta x} = 4 - 2 = \blue{2}\] The vertical change #\green{\Delta y}# is: \[\green{\Delta y} = f(4)-f(2)=6 - 2 = \green{4}\] So the average change is:

\[\frac{\green{\Delta y}}{\blue{\Delta x}} = \frac{\green{4}}{\blue{2}}=2\]

Note that the notation #[a,b]# for an interval can also be used for coordinates.

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Physical application

This works in the same way when we want to calculate the average speed of an airplane. Imagine an airplane which covers #4800# meters in #20# seconds. The average speed of the plane would be #\frac{4800}{20}=240# meters per second.

Linear functions

For linear functions, as in the example, the average change on each interval is the same. Suppose we have the function#f(x)=px+q#. The average change on each random interval #[a,b]# is given by
\[
\frac{f(b)-f(a)}{b-a}=\frac{p\cdot b + q - (p\cdot a +q)}{b-a}=\frac{p\cdot b + q - p\cdot a -q}{b-a}=\frac{p\cdot(a-b)}{a-b}=p
\]

We also call the average change on an interval the difference quotient.

Difference quotient

The difference quotient of a function #f# on an interval #[a,b]# is given by:

\[\dfrac{\green{\Delta y}}{\blue{\Delta x}}=\dfrac{\green{f(b)-f(a)}}{\blue{b-a}}\]

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Etymology

Literally quotient means 'the result of division'. The words 'difference' and 'quotient' put together convey the right essence: dividing one difference by another.

Consider the function #f(x)=x^3 + 2 x + 4#. What is the difference quotient on the interval #[1,2]#? Simplify your answer as much as possible.

#\frac{\Delta y}{\Delta x}=# #9#

#\begin{array}{rcl}\dfrac{\Delta y}{\Delta x}&=&\dfrac{f(2)-f(1)}{2-1}\\&&\phantom{xxx}\blue{a=1 \text{ and }b= 2}\\
&=&\dfrac{(1\cdot 2^3+2\cdot 2 + 4)-(1\cdot1^3+2\cdot1 + 4)}{2-1}\\&&\phantom{xxx}\blue{x=1 \text{ and } x=2 \text{ substituted in } f}\\ &=& \dfrac{9}{1}\\&&\phantom{xxx}\blue{\text{added}}\\ &=&9\\&&\phantom{xxx}\blue{\text{divided}}\end{array}#

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