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Interpreting, estimating, and using the derivative

A powerful feature of mathematics is that it can be studied both as an abstract discipline and as an applied one.

Introduction

A powerful feature of mathematics is that it can be studied both as an abstract discipline and as an applied one. For instance, calculus can be developed almost entirely as an abstract collection of ideas that focus on properties of functions. At the same time, if we consider functions that represent meaningful processes, calculus can describe our experience of physical reality. We have already learned that for the position function \(y = s(t)\) of a ball being tossed straight up in the air, the derivative of the position function, \(v(t) = s'(t)\), gives the ball's velocity at time \(t\).

In this section, we investigate several functions with contextual physical meaning and consider how the units on the independent variable, dependent variable, and the derivative function add to our understanding.

Exploration

Units of the derivative function

We know that the derivative of the function \(f\) at a fixed value \(x\) is given by \[\begin{aligned}\end{aligned}\], and that this value has several different interpretations. For instance, if we set \(x = a\), \(f'(a)\) is the slope of the tangent line to the graph of \(y = f(x)\) at the point \((a,f(a))\).

We sometimes write \(\frac{df}{dx}\) or \(\frac{dy}{dx}\) instead of \(f'(x)\), and these alternate notations emphasize the meaning of the derivative as the instantaneous rate of change of \(f\) with respect to \(x\). To understand the units on the derivative function, we use the fact that the instantaneous rate of change is the limit of the average rate of change. Recall that the average rate of change of \(f\) on the interval \([a,a+h]\) is \[\begin{aligned}\end{aligned}\], which is the quotient \[\begin{aligned}\end{aligned}\]. It follows that the units on the average rate of change are:

units of \(f\) per unit of \(x\).

Note that since \((a+h) - a = h\), we usually write \[\begin{aligned}\end{aligned}\].

Thus, when we take the limit of the average rate of change as \(h\) goes to zero, the derivative \[\begin{aligned}\end{aligned}\] has the same units: units of \(f\) per unit of \(x\). Said slightly differently, the units on the derivative function are units of output per unit of input for the variables and context of the original function, \(f\).

Example: Population growth

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 the year 2020). What does \[\begin{aligned}\end{aligned}\] mean in context?

Solution

The value \(P'(2) = 21.37\) tells us that the instantaneous rate of change of the city's population with respect to time at the start of 2022 is \(21.37\) thousand people per year, since \(P\) is measured in thousands of people and \(t\) is measured in years. In addition, because the derivative value tells us how fast the population is changing, we can also infer about how much the population will change: we expect that in the year from 2022 to 2023, about \(21,370\) people will be added to the city's population.

Condensed — the full section is in Boelkins, Active Calculus.

Toward more accurate derivative estimates

Recall that to estimate the value of \(f'(a)\) at a given value of \(x = a\), we can compute a difference quotient \(\frac{f(a+h)-f(a)}{h}\) with a relatively small value of \(h\). We usually 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. In the setting where we have limited data, we introduce the notion of a central difference for estimating the value of \(f'(a)\).

Example

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

Solution

We know that \(f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}\). 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, nor can we use smaller and smaller values of \(h\) to estimate the limit.

Instead, we have just two choices: using \(h = -1\) or \(h = 1\), depending on which point we pair with \((2,3.25)\).

So, one estimate is \[\begin{aligned}\end{aligned}\].

The other is \[\begin{aligned}\end{aligned}\].

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 \(x=2\). Doing so, we find that \[\begin{aligned}\end{aligned}\].

When we have two function values for \(f\) at equally spaced locations on opposite sides of \(x=a\), the approach of averaging the two estimates as we did in Example results in the best possible estimate we can find for \(f'(a)\) using the two known function values.

Observe that the central difference is also the average rate of change of \(f\) on \([a-h,a+h]\), \[\begin{aligned}\end{aligned}\], and thus we can even interpret the central difference as a slope of a certain secant line.

In Figure, we see a function that passes through the three data points given in Example: \((1,2.5)\), \((2,3.25)\), and \((3,3.265)\). The top line in the figure is the tangent line to the curve at \((2,f(2))\); for the function shown, the slope of that tangent line turns out to be \(f'(2) = 0.5199\). We found in Example that the central difference approximation is \(0.5625\), which is the slope of the secant line through the points \((1,f(1))\) and \((3,f(3))\) in Figure.

Condensed — the full section is in Boelkins, Active Calculus.

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. It also predicts about how much change we can expect when the input variable changes a small amount.
The units on the derivative function \(y = f'(x)\) are units of \(f\) per unit of \(x\). Again, this measures how fast the output of the function \(f\) changes when the input variable changes.
The central difference approximation to the value of the first derivative is given by \[\begin{aligned}\end{aligned}\]. This quantity measures the slope of the secant line to \(y = f(x)\) through the points \((a-h, f(a-h))\) and \((a+h, f(a+h))\). The central difference generates a good approximation of the derivative's value at \(x = a\).

Practice (9)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Wind turbines are used to generate electricity. The power, \(P\), measured in watts, depends on the windspeed, \(s\), measured in mph. For a particular wind turbine, \[\begin{aligned}\end{aligned}\] Find the average rate of change of \(P\) on the interval \([20, 22]\), \(AV_{[20, 22]}\), and write a sentence that explains the meaning of the value \(AV_{[20, 22]}\) to someone not taking calculus. Include units in your sentence.
A cup of coffee has its temperature \(F\) (in degrees Fahrenheit) at time \(t\) given by the function \(F(t) = 75 + 110 e^{-0.05t}\), where time is measured in minutes.
1. Use a central difference with \(h = 0.01\) to estimate the value of \(F'(10)\).
2. What are the units on the value of \(F'(10)\) that you computed in (a)? What is the practical meaning of the value of \(F'(10)\)?
3. Which do you expect to be greater: \(F'(10)\) or \(F'(20)\)? Why?
4. Write a sentence that describes the behavior of the function \(y = F'(t)\) on the time interval \(0 \le t \le 30\). How do you think its graph will look? Why?
បង្ហាញចម្លើយ
1. Using the central difference formula, we find that \[\begin{aligned}\end{aligned}\].
2. Since \(F\) is measured in degrees Fahrenheit and \(t\) in minutes, we have \(F'(10) \approx -3.33592\) degrees F per minute. This means that at the instant \(t=10\), the temperature of the coffee is decreasing at an instantaneous rate of about \(3.33592\) degrees per minute. Said differently, in the next minute, we'd expect the temperature of the coffee to drop about \(3.33592\) degrees.
3. We expect the coffee's temperature to decrease quickly early on when the difference between its temperature and the surroundings is greatest, and as time goes on for the coffee's temperature to approach that of the surrounding room. Keeping in mind that \(F\) is an always decreasing function so that \(F'\) is always negative, we observe that \(F'(10)\) should be less (more negative) than \(F'(20)\). These observations can also be confirmed by plotting a graph of \(F\).
4. We have noted that \(F'\) is always negative, and that we expect \(F'\) to get closer and closer to \(0\) as time goes on (that is, \(F'\) is increasing). As such, we expect \(F'\) to look something like the graph of \(y = -e^{-x}\). This is reasonable since \(-e^{-x}\) is also an exponential function, is always negative, is always increasing, and approaches \(0\) as \(x\) increases.
The temperature change \(T\) (in Fahrenheit degrees), in a patient, that is generated by a dose \(q\) (in milliliters), of a drug, is given by the function \(T = f(q)\).
1. What does it mean to say \(f(50) = 0.75\)? Write a complete sentence to explain, using correct units.
2. A person's sensitivity, \(s\), to the drug is defined by the function \(s(q) = f'(q)\). What are the units of sensitivity?
3. Suppose that \(f'(50) = -0.02\). Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a).
បង្ហាញចម្លើយ
1. To say that \(f(50) = 0.75\) means that if a patient takes a dose of \(50\) ml of a drug, the patient will experience a body temperature change of \(0.75\) degrees F.
2. Since the units of \(f\) are degrees F and the units of \(q\) are ml, it follows that the units of \(s(q) = f'(q)\) are degrees Fahrenheit per milliliter.
3. If \(f'(50) = -0.02\) degrees Fahrenheit per milliliter, this means that the instantaneous rate of change of the temperature change function is \(-0.02\) degrees F per ml at the dose-level of \(50\) ml. This means that for a patient taking a \(50\) ml dose, adding one more ml to the dose leads us to expect a temperature change that is about \(0.02\) degrees less than the temperature change induced by a \(50\) ml dose.
The velocity of a ball that has been tossed vertically in the air is given by \(v(t) = 16 - 32t\), where \(v\) is measured in feet per second, and \(t\) is measured in seconds. The ball is in the air from \(t = 0\) until \(t = 2\).
1. When is the ball's velocity greatest?
2. Determine the value of \(v'(1)\). Justify your thinking.
3. What are the units on the value of \(v'(1)\)? What does this value and the corresponding units tell you about the behavior of the ball at time \(t = 1\)?
4. What is the physical meaning of the function \(v'(t)\)?
បង្ហាញចម្លើយ
1. Observe that \(v(t) = 16-32t\) is a linear function with negative slope \(-32\), and thus decreasing. Since the function is of interest from \(t = 0\) to \(t=2\), the velocity is greatest at the left endpoint, when \(t=0\). This greatest velocity is \(v(0) = 16\) feet per second.
2. Since \(v\) is a linear function, its slope is constant, and hence the slope of every tangent line to the function is the same as the slope of the line the function generates. Thus, \(v'(1) = -32\).
3. Because \(v\) is measured in feet per second and \(t\) in seconds, the units on \(v'(1)\) are feet per second per second. Knowing that \(v'(1) = -32\), this tells us that the ball's velocity is decreasing at a rate of 32 feet per second per second; that is, for whatever the ball's velocity is at \(t = 1\), we would expect the velocity to decrease by (exactly) \(32\) feet per second over the next one second. (Normally this total decrease would be approximate, not exact, but since the velocity function is linear, we can compute the exact total change.)
4. The instantaneous rate of change of velocity is precisely the acceleration of the ball. Indeed, \(v'(t)\) tells us how the ball's velocity changes, and that it does so at a constant rate of \(-32\) feet per second per second. This is also consistent with the fact that acceleration due to gravity is constant.
The value, \(V\), of a particular automobile (in dollars) depends on the number of miles, \(m\), the car has been driven, according to the function \(V = h(m)\).
1. Suppose that \(h(40000) = 15500\) and \(h(55000) = 13200\). What is the average rate of change of \(h\) on the interval \([40000,55000]\), and what are the units on this value?
2. In addition to the information given in (a), say that \(h(70000) = 11100\). Determine the best possible estimate of \(h'(55000)\) and write one sentence to explain the meaning of your result, including units on your answer.
3. Which value do you expect to be greater: \(h'(30000)\) or \(h'(80000)\)? Why?
4. Write a sentence to describe the long-term behavior of the function \(V = h(m)\), plus another sentence to describe the long-term behavior of \(h'(m)\). Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing.
បង្ហាញចម្លើយ
1. To find the average rate of change of \(h\) on the interval \([40000,55000]\), we divide the change in \(h\) values (in dollars) on this interval by the change in \(m\) values (in miles) to obtain \[\begin{aligned}\end{aligned}\] dollars per mile. This tells us that from the mileage of \(40 000\) to \(50 000\) miles, on average the car's value decreases by \(0.153\) dollars per mile driven.
2. Since \(40 000\) and \(70 0000\) are equidistant from \(55 000\), we use the central difference quotient to approximate the derivative of \(h\) at \(55 0000\) to obtain \[\begin{aligned}\end{aligned}\] dollars per mile. This approximation tells us that at the instant the car has been driven \(55 000\) miles, the car's value is decreasing at an instantaneous rate of \(0.147\) dollars per mile. That is, during \(55 0001\)st mile, we'd expect the car's value to drop by \(0.147\) dollars.
3. As we pile up miles on our car, we expect the value to go down. However, we should expect the depreciation to be more extreme at first, but less in magnitude as we drive more miles and our car gets older. Note, though, the the value of the car is always decreasing and \(h'(m)\) is always negative. Thus, we should expect that the rate of change of \(h\) is increasing (becoming less negative) as \(m\) increases and that \(h'(30000) \lt h'(80000)\).
4. The value of a car is never negative. So we know that \(h(m) \geq 0\) for all \(m\). A new car decreases in value very quickly at first, and as the car gets older with more mileage, its value will depreciate to near zero. So \(h\) is a decreasing function. Since the value of our car always decreases, it follows that \(h'(m) \lt 0\) for all \(m \gt 0\). The fact that the depreciation slows as \(m\) increases implies that \(h'\) is an increasing (getting less negative) function. Since \(h'\) is increasing, it follows that \(h\) is bends upward; we will study this issue more in Section 1.6. So the graph of \(h\) might have the general shape of the graph of \(y = e^{-x}\) for positive values of \(x\).
A cup of coffee has its temperature \(F\) (in degrees Fahrenheit) at time \(t\) given by the function \(F(t) = 75 + 110 e^{-0.05t}\), where time is measured in minutes.
1. Use a central difference with \(h = 0.01\) to estimate the value of \(F'(10)\).
2. What are the units on the value of \(F'(10)\) that you computed in (a)? What is the practical meaning of the value of \(F'(10)\)?
3. Which do you expect to be greater: \(F'(10)\) or \(F'(20)\)? Why?
4. Write a sentence that describes the behavior of the function \(y = F'(t)\) on the time interval \(0 \le t \le 30\). How do you think its graph will look? Why?
បង្ហាញចម្លើយ
1. Using the central difference formula, we find that \[\begin{aligned}\end{aligned}\].
2. Since \(F\) is measured in degrees Fahrenheit and \(t\) in minutes, we have \(F'(10) \approx -3.33592\) degrees F per minute. This means that at the instant \(t=10\), the temperature of the coffee is decreasing at an instantaneous rate of about \(3.33592\) degrees per minute. Said differently, in the next minute, we'd expect the temperature of the coffee to drop about \(3.33592\) degrees.
3. We expect the coffee's temperature to decrease quickly early on when the difference between its temperature and the surroundings is greatest, and as time goes on for the coffee's temperature to approach that of the surrounding room. Keeping in mind that \(F\) is an always decreasing function so that \(F'\) is always negative, we observe that \(F'(10)\) should be less (more negative) than \(F'(20)\). These observations can also be confirmed by plotting a graph of \(F\).
4. We have noted that \(F'\) is always negative, and that we expect \(F'\) to get closer and closer to \(0\) as time goes on (that is, \(F'\) is increasing). As such, we expect \(F'\) to look something like the graph of \(y = -e^{-x}\). This is reasonable since \(-e^{-x}\) is also an exponential function, is always negative, is always increasing, and approaches \(0\) as \(x\) increases.
The temperature change \(T\) (in Fahrenheit degrees), in a patient, that is generated by a dose \(q\) (in milliliters), of a drug, is given by the function \(T = f(q)\).
1. What does it mean to say \(f(50) = 0.75\)? Write a complete sentence to explain, using correct units.
2. A person's sensitivity, \(s\), to the drug is defined by the function \(s(q) = f'(q)\). What are the units of sensitivity?
3. Suppose that \(f'(50) = -0.02\). Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a).
បង្ហាញចម្លើយ
1. To say that \(f(50) = 0.75\) means that if a patient takes a dose of \(50\) ml of a drug, the patient will experience a body temperature change of \(0.75\) degrees F.
2. Since the units of \(f\) are degrees F and the units of \(q\) are ml, it follows that the units of \(s(q) = f'(q)\) are degrees Fahrenheit per milliliter.
3. If \(f'(50) = -0.02\) degrees Fahrenheit per milliliter, this means that the instantaneous rate of change of the temperature change function is \(-0.02\) degrees F per ml at the dose-level of \(50\) ml. This means that for a patient taking a \(50\) ml dose, adding one more ml to the dose leads us to expect a temperature change that is about \(0.02\) degrees less than the temperature change induced by a \(50\) ml dose.
The velocity of a ball that has been tossed vertically in the air is given by \(v(t) = 16 - 32t\), where \(v\) is measured in feet per second, and \(t\) is measured in seconds. The ball is in the air from \(t = 0\) until \(t = 2\).
1. When is the ball's velocity greatest?
2. Determine the value of \(v'(1)\). Justify your thinking.
3. What are the units on the value of \(v'(1)\)? What does this value and the corresponding units tell you about the behavior of the ball at time \(t = 1\)?
4. What is the physical meaning of the function \(v'(t)\)?
បង្ហាញចម្លើយ
1. Observe that \(v(t) = 16-32t\) is a linear function with negative slope \(-32\), and thus decreasing. Since the function is of interest from \(t = 0\) to \(t=2\), the velocity is greatest at the left endpoint, when \(t=0\). This greatest velocity is \(v(0) = 16\) feet per second.
2. Since \(v\) is a linear function, its slope is constant, and hence the slope of every tangent line to the function is the same as the slope of the line the function generates. Thus, \(v'(1) = -32\).
3. Because \(v\) is measured in feet per second and \(t\) in seconds, the units on \(v'(1)\) are feet per second per second. Knowing that \(v'(1) = -32\), this tells us that the ball's velocity is decreasing at a rate of 32 feet per second per second; that is, for whatever the ball's velocity is at \(t = 1\), we would expect the velocity to decrease by (exactly) \(32\) feet per second over the next one second. (Normally this total decrease would be approximate, not exact, but since the velocity function is linear, we can compute the exact total change.)
4. The instantaneous rate of change of velocity is precisely the acceleration of the ball. Indeed, \(v'(t)\) tells us how the ball's velocity changes, and that it does so at a constant rate of \(-32\) feet per second per second. This is also consistent with the fact that acceleration due to gravity is constant.
The value, \(V\), of a particular automobile (in dollars) depends on the number of miles, \(m\), the car has been driven, according to the function \(V = h(m)\).
1. Suppose that \(h(40000) = 15500\) and \(h(55000) = 13200\). What is the average rate of change of \(h\) on the interval \([40000,55000]\), and what are the units on this value?
2. In addition to the information given in (a), say that \(h(70000) = 11100\). Determine the best possible estimate of \(h'(55000)\) and write one sentence to explain the meaning of your result, including units on your answer.
3. Which value do you expect to be greater: \(h'(30000)\) or \(h'(80000)\)? Why?
4. Write a sentence to describe the long-term behavior of the function \(V = h(m)\), plus another sentence to describe the long-term behavior of \(h'(m)\). Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing.
បង្ហាញចម្លើយ
1. To find the average rate of change of \(h\) on the interval \([40000,55000]\), we divide the change in \(h\) values (in dollars) on this interval by the change in \(m\) values (in miles) to obtain \[\begin{aligned}\end{aligned}\] dollars per mile. This tells us that from the mileage of \(40 000\) to \(50 000\) miles, on average the car's value decreases by \(0.153\) dollars per mile driven.
2. Since \(40 000\) and \(70 0000\) are equidistant from \(55 000\), we use the central difference quotient to approximate the derivative of \(h\) at \(55 0000\) to obtain \[\begin{aligned}\end{aligned}\] dollars per mile. This approximation tells us that at the instant the car has been driven \(55 000\) miles, the car's value is decreasing at an instantaneous rate of \(0.147\) dollars per mile. That is, during \(55 0001\)st mile, we'd expect the car's value to drop by \(0.147\) dollars.
3. As we pile up miles on our car, we expect the value to go down. However, we should expect the depreciation to be more extreme at first, but less in magnitude as we drive more miles and our car gets older. Note, though, the the value of the car is always decreasing and \(h'(m)\) is always negative. Thus, we should expect that the rate of change of \(h\) is increasing (becoming less negative) as \(m\) increases and that \(h'(30000) \lt h'(80000)\).
4. The value of a car is never negative. So we know that \(h(m) \geq 0\) for all \(m\). A new car decreases in value very quickly at first, and as the car gets older with more mileage, its value will depreciate to near zero. So \(h\) is a decreasing function. Since the value of our car always decreases, it follows that \(h'(m) \lt 0\) for all \(m \gt 0\). The fact that the depreciation slows as \(m\) increases implies that \(h'\) is an increasing (getting less negative) function. Since \(h'\) is increasing, it follows that \(h\) is bends upward; we will study this issue more in Section 1.6. So the graph of \(h\) might have the general shape of the graph of \(y = e^{-x}\) for positive values of \(x\).

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

Euler's number
2.71828…, the base whose exponential is its own derivative.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Interpreting, estimating, and using the derivative

In contexts other than the position of a moving object, what does the derivative of a function measure?
What are the units on the derivative function f', and how are they related to the units of the original function f?
What is a central difference, and how can one be used to estimate the value of the derivative at a point from given function data?
Given the value of the derivative of a function at a point, what can we infer about how the value of the function changes nearby?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

ព្យាយាមរបស់អ្នកផ្ទាល់

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

បន្ថែមទៀតក្នុង Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests