### Rules of differentiation: Applications of derivatives

### Elasticity

It is now clear that the derivative of a function is a measure for the absolute (instantaneous) rate of change of the function. But you can also consider the relative (proportional) rate of change. This is often used in economic analysis like: the Gross Domestic Product (GDP) of Greece increased by 0.5%. Especially in issues of pricing, we are interested in the relative (percentual) reduction in demand in relation to the relative (percentual) increase in price. In general, we define the relative change of a function as follows.

Elasticity

For a positive differentiable function #f(x)# the **elasticity**, or** relative rate of change**, for #x \gt 0# is defined as:

\[{\rm El}_x f(x)=f'(x) \cdot \dfrac{x}{f(x)}\tiny.\]

In order to understand this definition, consider the relative rate of change from #x# to #x+h# for some small number #h#: \[\begin{array}{rcl}\displaystyle \frac{\phantom{x}\frac{ f(x+h)-f(x)}{f(x)}\phantom{x}}{\frac{h}{x}} &=&{\dfrac{f(x+h)-f(x)}{h}}\cdot {\dfrac{x}{f(x)} }\tiny.\end{array}\]Since #\frac{f(x+h)-f(x)}{h}# is the difference quotient, it approaches #f'(x)# as #h\to0#. As a consequence, the relative rate of change from #x# to #x+h# tends to the limit #f'(x)\cdot\frac{x}{f(x)}# for #h\to0#.

The elasticity is a function #{\rm El}_x f# which only depends on #f#, and not on the argument #x#. The notation #{\rm El}_x f(x)# is somewhat confusing because the variable #x# is not determined by #f#. The reason for this choice is that later on we will work with functions of more variables and the index #x# then indicates with respect to which variable we are defining the elasticity.

We will illustrate this basic economic concept of *elasticity* by means of an example.

The skipper of the ferry is considering an increase of the rate, because he hopes to increase his income. Possibly some motorists will now stay away from the ferry because of the increased costs. Therefore, the skipper would like to compare changes in the variables associated with #p# and #q#.

How can he do that?

For calculating the result of the increase in the rate by 0.25 pounds to the demand, the skipper can use the difference quotient:

\[ \begin{array}{rcl}\frac{\Delta q}{\Delta p}&=& \dfrac{d(2.75)-d (2.50)}{0.25}\\&=&\dfrac{100\cdot (2.75)^2-8\cdot \cdot(2.75)+16-100\cdot (2.5)^2-8\cdot \cdot(2.5)+16}{0.25}\\&=&-275.00 \end{array}\] In fact, the difference quotient that is used here is the quotient of two absolute changes.

For a clearer picture of the dependence of demand on the price, working with relative changes is a better option. For, by itself, an absolute increase in the price by 1 pounds does not provide much information. A price increase by 1 to 2 pounds is an increase by 100%, whereas the same increase at a price by 1000 pounds is only an increase by 0.1%.

Instead of the quotient of the absolute changes in demand #q# and price #p#, we will determine the quotient of the relative changes in the question and the relative change in price. Since we can write the relative changes in demand and prices as #\frac{\Delta q}{q}# and #\frac{\Delta p}{p}#, respectively, this ratio is equal to:

\[\begin{array}{rcl} \frac{\Delta q}{q}/\frac{\Delta p}{p} &=& \dfrac{d(2.75)-d(2.50)}{d(2.50)}/\dfrac{0.25}{2.50} \\&=&\dfrac{100\cdot (2.75)^2-8\cdot \cdot(2.75)+16-100\cdot (2.5)^2-8\cdot \cdot(2.5)+16}{100\cdot (2.5)^2-8\cdot \cdot(2.5)+16}/0.1 \\&=& -3.06 \end{array}\]

A price increase of #2.50# by #0.25# is equivalent to a relative price increase by #0.1# or a percentage price increase by #10\%#; this is the denominator of the ratio. It makes no difference to this formula if we take the quotient of relative change or percent change: when we use percentage change, we multiply both the numerator and denominator with #100#. Thus, in this case, a price increase by #10\%# leads to a decrease by #-31\%# of interest in using the ferry. This decrease is approximately three times larger than the gain.

We can take smaller differences for a more precise approach. To this end we rewrite the first quotient as follows:

Since #\frac{d(p+\Delta p)-d(p)}{\Delta p}# is the difference quotient, as #\Delta p \to 0# this approaches #d'(p)#. As a consequence, the best approach is the limit: \[d'(p)\cdot\frac{p}{d(p)}=\left(2\cdot p-8\right)\cdot\dfrac{p}{p^2-8\cdot p+16}={{2\cdot p}\over{p-4}}\tiny.\] Substituting #p=2.50# in this expression, we find the relative decline of interest in using the ferry to be #3.33#.

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