The Derivative as a Rate of Change

How do you measure how fast one quantity changes with another?

You take the derivative of the first quantity with respect to the second, and read it in units of the first per unit of the second. If y depends on x, the difference quotient (\Delta y)/(\Delta x) is the average rate of change over a step, and its limit (dy)/(dx) is the instantaneous rate of change. The word "instantaneous" is used even when x is not time, and it is often dropped: "the rate of change" means the instantaneous one. This is the same limit as the derivative at a point, now read as a rate.

The units come out of the quotient. If the area A of a circle is measured in square meters and its diameter D in meters, then (dA)/(dD) is in square meters per meter. For A=(π D^2)/(4), (dA)/(dD)=(π D)/(2), so when D=10 m the area is growing at 5π≈15.71 square meters for every extra meter of diameter.

The rate depends on the variable you choose. The same circle measured by its radius has A=π r^2 and (dA)/(dr)=2π r. At D=2r that is twice as large as (dA)/(dD), because the diameter grows twice as fast as the radius. Like kilometers per hour against meters per second, a rate always belongs to a pair of units.

The rates that turn out to be geometry. For a circle, (dA)/(dr)=2π r is its circumference; for a sphere, (dV)/(dr)=4π r^2 is its surface area; for a cube, (dV)/(da)=3a^2 is the area of three faces. The reason is the same in each: a slightly larger shape is the old one plus a thin layer, and the layer's volume is its area times its thickness.

Estimating a change. Since \Delta y≈(dy)/(dx) \Delta x for a small step, the rate gives a quick estimate. A sphere's radius grows from 2 m to 2.2 m: the volume rate at r=2 is 4π·4≈50.27 cubic meters per meter, so \Delta V≈50.27·0.2≈10.05 cubic meters. The exact change is \frac43π\left(2.2^3-2^3\right)≈11.09, about 10\% more: the estimate uses the slope at the start of the step, and the surface grows as the sphere does. A smaller step shrinks the error.

Pick a shape below and compare the exact increase with the rate.

How do velocity, speed and acceleration come from derivatives?

For a body moving along a line with position s(t), velocity is the derivative of position, acceleration is the derivative of velocity, and jerk is the derivative of acceleration: v=(ds)/(dt), a=(dv)/(dt)=(d^2s)/(dt^2) and j=(da)/(dt)=(d^3s)/(dt^3). The line is usually horizontal or vertical, and s is the coordinate on it.

• Displacement and average velocity. From t to t+\Delta t the displacement is \Delta s=s(t+\Delta t)-s(t) and the average velocity is (\Delta s)/(\Delta t).
• Velocity is the limit of the average velocity as \Delta t→0. Its sign is the direction: positive when s is increasing (forward, or upward on a vertical line), negative when s is decreasing.
• Speed is the absolute value |v|, the rate of progress with no direction. A car driven to a friend's house and back at 50 km/h has a speedometer that shows 50 both ways, though its velocity is +50 then -50.
• Acceleration says how fast the velocity changes: it is the second derivative of position. It is measured in units of velocity per unit of time, such as m/s^2.
• Jerk is the rate of change of acceleration. A ride feels jerky when the acceleration changes abruptly, not necessarily when it is large.

A graph of s against t is not the path of the body: the path lies on the line, and the graph shows where the body is when. Its slope is the velocity.

Speeding up or slowing down. A body speeds up when velocity and acceleration have the same sign (it picks up speed in the direction it is going) and slows down when they have opposite signs. Acceleration alone does not tell you: a<0 with v<0 is a body moving backward and speeding up.

An example. Take s=t^3-6t^2+9t meters for 0\le t\le4 seconds. Then v=3t^2-12t+9=3(t-1)(t-3) and a=6t-12. The velocity is zero at t=1 and t=3, where the body turns round: s(1)=4 and s(3)=0. The table of the motion is

• 0<t<1: v>0, a<0, forward and slowing down;
• 1<t<2: v<0, a<0, backward and speeding up;
• 2<t<3: v<0, a>0, backward and slowing down;
• 3<t<4: v>0, a>0, forward and speeding up.

The displacement over [0,4] is s(4)-s(0)=4 m and the average velocity is 1 m/s, but the body actually travels 4+4+4=12 m, because the distance is added up piece by piece between the turns. At t=2, where a=0, the speed is |v(2)|=3 m/s.

Type your own position function below: it finds the turns exactly (even when they are not fractions), classifies every interval, and adds up the distance.

How do you analyze free fall and a thrown object?

Near the surface of the Earth every falling body has the same constant acceleration g≈9.8 m/s^2, so its distance fallen from rest is s=\frac12gt^2=4.9t^2 meters, its velocity is v=9.8t and its acceleration is the constant 9.8. The formula comes from Galileo's experiments and holds in a vacuum; it models a dense, heavy object well for the first few seconds, before air resistance matters.

A dropped ball bearing. In the first 3 s it falls s(3)=4.9·9=44.1 m. Its velocity at that time is v(3)=9.8·3=29.4 m/s downward, its speed is 29.4 m/s and its acceleration is 9.8 m/s^2. Because the acceleration is constant, its jerk is (d)/(dt)(9.8)=0: nothing about free fall is jerky.

A rock thrown straight up. A blast sends a rock up with launch speed 49 m/s, so its height is s=49t-4.9t^2 meters, measured upward, and v=49-9.8t.

• How high does it go? At the top the velocity is 0: 49-9.8t=0 gives t=5 s and s(5)=245-122.5=122.5 m.
• Its speed at 78.4 m, going up and coming down. Solving 49t-4.9t^2=78.4 gives 4.9(t^2-10t+16)=0, which factors as (t-2)(t-8)=0: the rock is at 78.4 m after 2 s going up and after 8 s coming down. The velocities are v(2)=29.4 and v(8)=-29.4 m/s: the same speed, in opposite directions.
• Its acceleration. a=(d)/(dt)(49-9.8t)=-9.8 m/s^2 the whole time, up and down: gravity pulls downward, so the rock slows on the way up and speeds up on the way down. At the very top the velocity is zero and the acceleration is still -9.8.
• When does it land? s=49t-4.9t^2=4.9 t (10-t)=0 gives t=0 (the blast) and t=10 s. Its velocity on landing is v(10)=-49 m/s: it comes back at the speed it left with.

The same story with any launch speed v_0. From s=v_0t-\frac12gt^2 the top is at t=(v_0)/(g), its height is (v_0^2)/(2g), the landing is at t=(2v_0)/(g) (twice as long as the way up) and the landing velocity is -v_0. With v_0=49 and g=9.8 those are 5 s, (2401)/(19.6)=122.5 m and 10 s.

The graph of s against t is a parabola, but the rock moves along a vertical line: the parabola shows height against time, and its slope at each instant is the velocity. The tool of the previous section handles these: try the preset 49t - 4.9t^2 on 0\le t\le10.

What are marginal cost, marginal revenue and marginal profit?

In economics a marginal quantity is a derivative: the marginal cost is the rate of change of the cost c(x) with respect to the number of units x, (dc)/(dx), and it estimates what one more unit costs. The average cost of the next h units is (c(x+h)-c(x))/(h); its limit as h→0 is the marginal cost. Loosely, the marginal cost is defined as the extra cost of one more unit, c(x+1)-c(x), which is approximated by c'(x) when the slope of c does not change quickly near x. The approximation often works well for large x.

Cost, fixed and variable. A total cost function is often a cubic polynomial c(x)=α x^3+β x^2+\gamma x+\delta. The constant \delta is the fixed cost (rent, equipment, management) that does not depend on x; the other terms are variable costs (materials, labor). A cubic is usually enough to capture how cost behaves over a realistic range.

An example. Let the cost of making x radiators a day be c(x)=x^3-6x^2+15x dollars, and the revenue from selling them r(x)=x^3-3x^2+12x dollars. The shop makes 10 a day. What does an eleventh cost, and what does it bring?

• Marginal cost: c'(x)=3x^2-12x+15, so c'(10)=300-120+15=195 dollars.
• Marginal revenue: r'(x)=3x^2-6x+12, so r'(10)=300-60+12=252 dollars.
• Estimated profit from the extra radiator: 252-195=57 dollars.

The exact figures are c(11)-c(10)=770-550=220 dollars and r(11)-r(10)=1100-820=280 dollars, a profit of 60. The estimates 195, 252 and 57 are close but not equal, because the cost and revenue slopes rise as the output grows: the derivative is the rate at x=10, and one whole unit is a big step. A smaller step would agree better.

Marginal profit. Profit is revenue minus cost, p=r-c, so marginal profit is r'-c'. It is zero where marginal revenue equals marginal cost, a place where producing a little more or a little less changes the profit by about nothing. Finding the best production level among such places is a question about optimization, taken up later.

Marginal tax rate. Marginal rates appear in taxes too. If your marginal income tax rate is 28\% and your income rises by 1000 dollars, you pay about 280 dollars more in tax. That does not mean you pay 28\% of your whole income: it means (dT)/(dI)=0.28, the tax on each extra dollar. As income rises you may enter a higher bracket, and the marginal rate rises.

Type any cost and revenue below and compare each marginal value with the exact one-unit change.

How does the derivative measure sensitivity to change?

A function is sensitive to its input where a small change in x produces a large change in f(x), and the derivative f'(x) measures exactly that: the larger |f'(x)|, the more sensitive f is at x. On a graph that is the steepness of the tangent line.

Peas and genes. Gregor Mendel found that if p (between 0 and 1) is the frequency of the gene for smooth skin in peas and 1-p that of wrinkled skin, then the proportion of smooth peas in the next generation is y=2p(1-p)+p^2=2p-p^2. The derivative is (dy)/(dp)=2-2p, close to 2 for p near 0 and close to 0 for p near 1. So the result is much more sensitive to p when p is small: adding a few smooth-skin genes to a population that is mostly wrinkled has a much bigger effect on later generations than adding the same number to a population that is already mostly smooth.

Braking distance. A model of the total stopping distance of a car at speed V km/h is s=0.21V+0.0063V^2 meters: the linear part is the distance covered while the driver reacts, the quadratic part the braking distance. The derivative is (ds)/(dV)=0.21+0.0126V, which is 0.84 meters per km/h at 50 km/h and 1.47 meters per km/h at 100 km/h. The stopping distance is much more sensitive to speed when you are already going fast: from 50 to 100 km/h the speed doubles but the stopping distance goes from 26.25 m to 84 m, more than three times as far.

How to read a sensitivity. The derivative is an absolute rate (units of f per unit of x), so it is the right measure when the units mean something to you: meters of distance for each km/h. A graph of f' next to the graph of f, as in graphing the derivative, shows at a glance where f is most and least sensitive.

How do you find rates of change from data in code?

When you have measurements instead of a formula, a derivative is a difference quotient of the data: use numpy.gradient on the positions to get velocities, and again on the velocities to get accelerations. The function uses central differences inside the data and one-sided differences at the two ends. For a quadratic like free fall the central difference is exact, so the interior velocities are exactly 9.8t and the interior accelerations 9.8, while the two end values are off, because a one-sided difference is only first-order accurate.

import numpy as np

t = np.linspace(0, 3, 7)               # a reading every half second
s = 4.9 * t**2                         # distance fallen, in meters
v = np.gradient(s, t)                  # velocity
a = np.gradient(v, t)                  # acceleration
print(np.round(v, 2))                  # [ 2.45  4.9   9.8  14.7  19.6  24.5  26.95]
print(np.round(9.8 * t, 2))            # [ 0.    4.9   9.8  14.7  19.6  24.5  29.4 ]   (the exact velocities)
print(np.round(a, 2))                  # [4.9  7.35 9.8  9.8  9.8  7.35 4.9 ]            (the exact value is 9.8)

c = lambda x: x**3 - 6*x**2 + 15*x     # the radiators' cost
print(c(11) - c(10), 3*10**2 - 12*10 + 15)   # 220 195: the next unit against the marginal cost

Differentiating data makes noise louder. Real measurements have errors, and a derivative divides a difference of two nearby readings by a small time step, so a tiny position error becomes a larger velocity error, and differentiating again makes it far larger. The run below adds random noise of about 0.01 m (a centimeter) to 301 position readings of the same fall, 0.01 s apart. The velocity from them is off by about 0.7 m/s, and the acceleration by about 60 m/s², which is six times the true value. To get usable rates from noisy data, smooth the data first, fit a model such as s=\frac12gt^2 to it, or use readings farther apart.

rng = np.random.default_rng(0)
t = np.linspace(0, 3, 301)                           # a reading every 0.01 s
s = 4.9 * t**2 + rng.normal(0, 0.01, t.size)         # the same fall, with 1 cm of noise
v = np.gradient(s, t)
a = np.gradient(v, t)
print(round(float(np.std(s - 4.9*t**2)), 4))              # 0.0102   position error, in m
print(round(float(np.std(v[2:-2] - 9.8*t[2:-2])), 3))     # 0.716    velocity error, in m/s
print(round(float(np.std(a[4:-4] - 9.8)), 1))             # 62.7     acceleration error, in m/s^2

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
  double t[7], s[7], v[7], a[7];                       // a reading every half second
  for (int i = 0; i < 7; i++) { t[i] = 0.5 * i; s[i] = 4.9 * t[i] * t[i]; }   // distance fallen, in meters
  v[0] = (s[1] - s[0]) / 0.5;                          // velocity: one-sided at the ends, central inside
  v[6] = (s[6] - s[5]) / 0.5;
  for (int i = 1; i < 6; i++) v[i] = (s[i + 1] - s[i - 1]) / 1.0;
  a[0] = (v[1] - v[0]) / 0.5;                          // acceleration: the same, applied to the velocity
  a[6] = (v[6] - v[5]) / 0.5;
  for (int i = 1; i < 6; i++) a[i] = (v[i + 1] - v[i - 1]) / 1.0;
  for (int i = 0; i < 7; i++) printf("%6.2f", v[i]);   // 2.45 4.9 9.8 14.7 19.6 24.5 26.95
  cout << endl;
  for (int i = 0; i < 7; i++) printf("%6.2f", 9.8 * t[i]);   // the exact velocities
  cout << endl;
  for (int i = 0; i < 7; i++) printf("%6.2f", a[i]);   // the exact acceleration is 9.8
  cout << endl;

  cout << (11 * 11 * 11 - 6 * 11 * 11 + 15 * 11) - (10 * 10 * 10 - 6 * 10 * 10 + 15 * 10) << " "
       << 3 * 10 * 10 - 12 * 10 + 15 << endl;          // 220 195: the next unit against the marginal cost

  // noise grows with every derivative: readings every 0.01 s, each off by 0, +0.01 or -0.01 m
  double noise[3] = {0.01, -0.01, 0};
  double e[40], ve[40], ae[40], h = 0.01;
  for (int i = 0; i < 40; i++) e[i] = noise[i % 3];
  for (int i = 1; i < 39; i++) ve[i] = (e[i + 1] - e[i - 1]) / (2 * h);
  double maxS = 0, maxV = 0, maxA = 0;
  for (int i = 0; i < 40; i++) if (e[i] > maxS) maxS = e[i];
  for (int i = 2; i < 38; i++) {
    ae[i] = (ve[i + 1] - ve[i - 1]) / (2 * h);
    if (ve[i] > maxV) maxV = ve[i];
    if (ae[i] > maxA) maxA = ae[i];
  }
  printf("%.2f %.1f %.1f\n", maxS, maxV, maxA);          // the error in m, in m/s and in m/s^2
  return 0;
}

/* output:
  2.45  4.90  9.80 14.70 19.60 24.50 26.95
  0.00  4.90  9.80 14.70 19.60 24.50 29.40
  4.90  7.35  9.80  9.80  9.80  7.35  4.90
220 195
0.01 1.0 75.0
*/

Who first measured instantaneous speed and marginal change?

Speed at an instant was a puzzle for two thousand years before calculus gave it a definition, and the economists' "margin" arrived two centuries after that.

• Galileo. In his Discourses on Two New Sciences (1638) Galileo showed that a body falling from rest covers distances in proportion to the square of the time, s\propto t^2, which is the law s=\frac12gt^2 used here. He found it by rolling balls down inclined planks, which slows the motion enough to time it.
• The Oxford Calculators. Scholars at Merton College, Oxford, in the 1330s distinguished uniform from non-uniform motion and defined "instantaneous velocity" in words; their mean speed theorem was drawn as a graph of speed against time by Nicole Oresme around 1350, centuries before the derivative.
• Newton. In the 1660s Isaac Newton treated velocity as the fluxion of position, the rate at which a flowing quantity changes, and acceleration as the fluxion of velocity: the derivative was invented, in part, to talk about motion.
• Marginal economics. Augustin Cournot used the derivative in his 1838 book on the mathematical theory of wealth to find the output that maximizes a firm's profit. The word "marginal" in economics comes from Friedrich von Wieser (1884), and the "marginal revolution" of the 1870s made it central.

What mistakes do people make with rates of change?

The most common mistakes with rates of change are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: the derivative as a rate of change at a glance

• Rate of change: (dy)/(dx)=\lim(\Delta y)/(\Delta x) in units of y per unit of x; it depends on the variable chosen; \Delta y≈(dy)/(dx)\Delta x for small steps.
• Motion: v=(ds)/(dt) (sign = direction), speed =|v|, a=(d^2s)/(dt^2), j=(da)/(dt); speeding up when v and a have the same sign; distance adds |\Delta s| between the turns, displacement is s(t_1)-s(t_0).
• Free fall: s=4.9t^2, v=9.8t, a=9.8, j=0; thrown up at v_0: top at (v_0)/(g), height (v_0^2)/(2g), landing at (2v_0)/(g) with velocity -v_0.
• Marginals: marginal cost c'(x)≈ c(x+1)-c(x), marginal revenue r', marginal profit r'-c'; marginal tax rate is the tax on the next unit of income.
• Sensitivity: a large |f'| means a small change in x makes a large change in f (Mendel's peas, braking distance).
• Data and code: numpy.gradient for velocity and acceleration from readings; each derivative amplifies noise.

Glossary: key terms

Velocity

The derivative of position with respect to time; its size is the speed and its sign is the direction of motion along a line.

Speed

The absolute value of velocity, the rate of progress regardless of direction, which is never negative.

Acceleration

The derivative of velocity with respect to time, which is also the second derivative of position; it measures how fast a body gains or loses velocity.

Jerk

The derivative of acceleration with respect to time, which is the third derivative of position; a ride is jerky when the acceleration changes abruptly.

Displacement

The change in position from the start to the end of a time interval, found by subtracting the starting position from the final one; it can be smaller than the distance travelled.

Marginal cost

The derivative of the cost of production with respect to the number of units produced, which approximates the cost of making one more unit.

Marginal revenue

The derivative of revenue with respect to the number of units sold, which approximates the extra revenue from selling one more unit.

Sensitivity to change

How strongly a function responds to a small change in its input, measured by the size of its derivative: the steeper the graph, the more sensitive the function.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 3.4: The Derivative as a Rate of Change. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 3.7: Rates of Change in the Natural and Social Sciences. Cengage Learning.
3. Galileo Galilei (1638). Discorsi e dimostrazioni matematiche intorno a due nuove scienze (Discourses on Two New Sciences). Elsevier, Leiden.
4. Augustin Cournot (1838). Recherches sur les principes mathématiques de la théorie des richesses. Hachette, Paris.
5. NumPy Developers. NumPy reference: numpy.gradient. numpy.org. https://numpy.org/doc/stable/reference/generated/numpy.gradient.html.

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