1.5: Interpretating, Estimating, and Using the Derivative

Units of the derivative function

As we now know, the derivative of the function $f$ at a fixed value $x$ is given by

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\text{,}$$

and this value has several different interpretations. If we set $x = a\text{,}$ one meaning of $f'(a)$ is the slope of the tangent line at the point $(a,f(a))\text{.}$

We also sometimes write $\frac{df}{dx}$ or $\frac{dy}{dx}$ instead of $f'(x)\text{,}$ and these alternate notations help us see the units (and thus the meaning) of the derivative as the instantaneous rate of change of $f$ with respect to $x$. The units on the slope of the secant line, $\frac{f(x+h)-f(x)}{h}\text{,}$ are “units of $y$ per unit of $x\text{,}$” and when we take the limit as $h$ goes to zero, the derivative $f'(x)$ has the same units: units of $y$ per unit of $x\text{.}$ It is helpful to remember that the units on the derivative function are “units of output per unit of input,” for the variables of the original function.

For example, suppose that the function $y = P(t)$ measures the population of a city (in thousands) at the start of year $t$ (where $t = 0$ corresponds to 2010 AD). We are told that $P'(2) = 21.37\text{.}$ What is the meaning of this value? Well, since $P$ is measured in thousands and $t$ is measured in years, we can say that the instantaneous rate of change of the city's population with respect to time at the start of 2012 is 21.37 thousand people per year. We therefore expect that in the coming year, about 21,370 people will be added to the city's population.

Toward more accurate derivative estimates

Recall that to estimate the value of $f'(x)$ at a given $x\text{,}$ we calculate a difference quotient $\frac{f(x+h)-f(x)}{h}$ with a relatively small value of $h\text{.}$ We should use both positive and negative values of $h$ in order to account for the behavior of the function on both sides of the point of interest. To that end, we introduce the notion of a central difference and its role in estimating derivatives.

Example $1.5.1$

Suppose that $y = f(x)$ is a function for which three values are known: $f(1) = 2.5\text{,}$ $f(2) = 3.25\text{,}$ and $f(3) = 3.625\text{.}$ Estimate $f'(2)\text{.}$

Answer

We know that $f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}\text{.}$ But since we don't have a graph or a formula for the function, we can neither sketch a tangent line nor evaluate the limit algebraically. We can't even use smaller and smaller values of $h$ to estimate the limit. Instead, we have just two choices: using $h = -1$ or $h = 1\text{,}$ depending on which point we pair with $(2,3.25)\text{.}$

So, one estimate is

$$f'(2) \approx \frac{f(1)-f(2)}{1-2} = \frac{2.5-3.25}{-1} = 0.75\text{.}$$

The other is

$$f'(2) \approx \frac{f(3)-f(2)}{3-2} = \frac{3.625-3.25}{1} = 0.375\text{.}$$

Because the first approximation looks backward from the point $(2,3.25)$ and the second approximation looks forward, it makes sense to average these two estimates in order to account for behavior on both sides of $x=2\text{.}$ Doing so, we find that

$$f'(2) \approx \frac{0.75 + 0.375}{2} = 0.5625\text{.}$$

The intuitive approach to average the two estimates found in Example $1.5.1$ is in fact the best possible way estimate to a derivative when we have just two function values for $f$ on opposite sides of the point of interest.

To see why, we think about the diagram in Figure $1.5.2$ . On the left, we see the two secant lines with slopes that come from computing the backward difference $\frac{f(1)-f(2)}{1-2} = 0.75$ and from the forward difference $\frac{f(3)-f(2)}{3-2} = 0.375\text{.}$ Note how the first slope over-estimates the slope of the tangent line at $(2,f(2))\text{,}$ while the second slope underestimates $f'(2)\text{.}$ On the right, we see the secant line whose slope is given by the central difference

Note that this central difference has the same value as the average of the forward and backward differences (and it is straightforward to explain why this always holds). The central difference yields a very good approximation to the derivative's value, because it yields a line closer to being parallel to the tangent line.

The central difference approximation to the value of the first derivative is given by

This quantity measures the slope of the secant line to $y = f(x)$ through the points $(a-h, f(a-h))$ and $(a+h, f(a+h))\text{.}$

Activity $1.5.2$

A potato is placed in an oven, and the potato's temperature $F$ (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time $t$ is measured in minutes.

Table $1.5.1$. Temperature data in degrees Fahrenheit.

$t$	$0$	$15$	$30$	$45$	$60$	$75$	$90$
$F(t)$	$70$	$180.5$	$251$	$296$	$324.5$	$342.8$	$354.5$

1. Use a central difference to estimate the instantaneous rate of change of the temperature of the potato at $t= 30\text{.}$ Include units on your answer.
2. Use a central difference to estimate the instantaneous rate of change of the temperature of the potato at $t= 60\text{.}$ Include units on your answer.
3. Without doing any calculation, which do you expect to be greater: $F'(75)$ or $F'(90)\text{?}$ Why?
4. Suppose it is given that $F(64) = 330.28$ and $F'(64) = 1.341\text{.}$ What are the units on these two quantities? What do you expect the temperature of the potato to be when $t = 65\text{?}$ when $t = 66\text{?}$ Why?
5. Write a couple of careful sentences that describe the behavior of the temperature of the potato on the time interval $[0,90]\text{,}$ as well as the behavior of the instantaneous rate of change of the temperature of the potato on the same time interval.

Activity $1.5.3$

A company manufactures rope, and the total cost of producing $r$ feet of rope is $C(r)$ dollars.

1. What does it mean to say that $C(2000) = 800\text{?}$
2. What are the units of $C'(r)\text{?}$
3. Suppose that $C(2000) = 800$ and $C'(2000) = 0.35\text{.}$ Estimate $C(2100)\text{,}$ and justify your estimate by writing at least one sentence that explains your thinking.
4. Do you think $C'(2000)$ is less than, equal to, or greater than $C'(3000)\text{?}$ Why?
5. Suppose someone claims that $C'(5000) = -0.1\text{.}$ What would the practical meaning of this derivative value tell you about the approximate cost of the next foot of rope? Is this possible? Why or why not?

Activity $1.5.4$

Researchers at a major car company have found a function that relates gasoline consumption to speed for a particular model of car. In particular, they have determined that the consumption $C\text{,}$ in liters per kilometer, at a given speed $s\text{,}$ is given by a function $C = f(s)\text{,}$ where $s$ is the car's speed in kilometers per hour.

1. Data provided by the car company tells us that $f(80) = 0.015\text{,}$ $f(90) = 0.02\text{,}$ and $f(100) = 0.027\text{.}$ Use this information to estimate the instantaneous rate of change of fuel consumption with respect to speed at $s = 90\text{.}$ Be as accurate as possible, use proper notation, and include units on your answer.
2. By writing a complete sentence, interpret the meaning (in the context of fuel consumption) of “$f(80) = 0.015\text{.}$”
3. Write at least one complete sentence that interprets the meaning of the value of $f'(90)$ that you estimated in (a).

In Section 1.4, we learned how use to the graph of a given function $f$ to plot the graph of its derivative, $f'\text{.}$ It is important to remember that when we do so, the scale and the units on the vertical axis often have to change to represent $f'\text{.}$ For example, suppose that $P(t) = 400-330e^{-0.03t}$ tells us the temperature in degrees Fahrenheit of a potato in an oven at time $t$ in minutes. In Figure $1.5.3$ , we sketch the graph of $P$ on the left and the graph of $P'$ on the right.

Notice that the vertical scales are different in size and different in units, as the units of $P$ are $^\circ$F, while those of $P'$ are $^\circ$F/min.

Summary

• The derivative of a given function $y=f(x)$ measures the instantaneous rate of change of the output variable with respect to the input variable.
• The units on the derivative function $y = f'(x)$ are units of $y$ per unit of $x\text{.}$ Again, this measures how fast the output of the function $f$ changes when the input of the function changes.
• The central difference approximation to the value of the first derivative is given by
  $$f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}\text{.}$$

  This quantity measures the slope of the secant line to $y = f(x)$ through the points $(a-h, f(a-h))$ and $(a+h, f(a+h))\text{.}$ The central difference generates a good approximation of the derivative's value.