The amount of change formula and the estimate it licenses, motion along a line with velocity, speed and acceleration and the rule for speeding up or slowing down, marginal cost, revenue and profit in economics, population growth rates, and the discipline of attaching correct units to every applied derivative.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
Derivatives as Rates of Change
Objectives
Five outcomes, and not one of them is a new differentiation rule. This section is about what the derivative means once you can compute it.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 230-240 — the section these objectives are drawn from
Warm-up
Section 3.3 made differentiation mechanical. A polynomial's derivative now takes one line, which raises a question the earlier sections postponed.
Discussion prompt
You compute that a cost function has derivative 40 at x equal to 500. What does the number 40 actually tell a manager, and what are its units?
Hint: A derivative is a limit of output changes divided by input changes.
Answer:
\[ C'(500) = 40 \quad \text{with units of dollars per unit} \]
It says that around a production level of 500, each additional unit costs about 40 dollars. Not that the total cost is 40, and not that costs are rising by 40 percent.
The units carry the whole interpretation: dollars divided by units is a cost PER UNIT, which is a rate. This section is about reading derivatives that way across physics, economics and biology — the skill Chapter 4 will assume throughout.
Concept
The derivative gives the rate at which one quantity changes per unit change in another. Multiplying that rate by a small change in the input estimates the resulting change in the output.
amount of change — If a quantity changes by a small amount in its input, the resulting change in the output is approximately the derivative times that input change. The approximation improves as the input change shrinks.
\[ f(a + h) - f(a) \approx f'(a)\,h \]
The approximation is exactly the tangent line standing in for the curve, which is why the error is whatever the curve does that a straight line cannot follow. Section 4.2 makes this precise and quantifies the error.
Figure (svg): A curve with its tangent, showing the derivative used to estimate the change over one unit
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 230-232
Section
Section 1
Concept
Rearranging the definition of the derivative gives an estimate: the change in output over a small input change is about the derivative times that change. For a one-unit change the estimate is just the derivative itself.
the estimate from a rate — The change in f over an interval of length h starting at a is approximately f prime of a times h. It is exact only when f is linear, and otherwise carries an error that grows with h.
\[ \Delta f \approx f'(a)\,\Delta x \]
The reason a one-unit change works at all is that in most applications the unit is small compared with the scale of the problem — one more item out of five hundred, one more second out of an hour.
Figure (svg): A curve with its tangent, showing the derivative used to estimate the change over one unit
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 230-233 — amount of change formula
Picture it
The estimate and the actual value, one unit along.
Figure (svg): A curve with its tangent, showing the derivative used to estimate the change over one unit
The estimate walks along the tangent line and the true value walks along the curve. Where the curve bends away, the gap opens — and for a curve bending upward the estimate is always an undershoot.
Worked example
Example 3.27. The rate, times the change.
\[ \text{For } f(x) = x^2, \text{ estimate } f(3.1) - f(3) \text{ and compare with the exact value.} \]
Differentiate
Why: Power rule.
\[ f'(x) = 2 x \]
Evaluate the rate at the starting point
Why: Twice 3.
\[ f'(3) = 6 \]
Multiply by the input change
Why: The change is 0.1.
\[ 6 \times 0.1 = 0.6 \]
Compute the exact change for comparison
Why: 9.61 minus 9.
\[ 0.61 \]
Compare
Why: The estimate is slightly low.
\[ \text{error } 0.01 \]
Figure (svg): The solution to Worked example estimating with a derivative shown as a ladder of expressions, one row per legal move
\[ \Delta f \approx 0.6, \qquad \text{exact } 0.61 \]
Verify: check what happens with a bigger step
Why: Over a full unit, from 3 to 4, the estimate is 6 and the exact change is 7 — an error of 1, a hundred times larger than before. The error grows roughly with the SQUARE of the step, which is why a tenth-sized step gave a hundredth-sized error. The estimate is always low here because the parabola bends upward and the tangent lies beneath it, and knowing the direction of the error is often as useful as its size.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 231-232
Fill the middle
The squaring function, stepping a tenth of a unit from 3.
Fill in the blanks
\Delta f \approx f'(3)\cdot 0.1 = 6 \cdot 0.1 = 0.6
Why: The estimate is 0.6 and the exact change is 0.61. The estimate is low because the parabola bends upward and the tangent lies beneath the curve.
Worked example
Checkpoint 3.27. The commonest applied form.
\[ \text{Cost is } C(x) = 1000 + 12x + 0.1x^2. \text{ Estimate the cost of the } 51\text{st unit.} \]
Differentiate
Why: Termwise.
\[ C'(x) = 12 + 0.2 x \]
Evaluate at the level before the extra unit
Why: At x equal to 50.
\[ C'(50) = 22 \]
Interpret
Why: Dollars per additional unit.
\[ \text{about } 22\text{ dollars} \]
Check against the exact difference
Why: C(51) minus C(50).
\[ 22.10 \]
Figure (svg): The solution to Worked example a one-unit estimate in context shown as a ladder of expressions, one row per legal move
\[ C'(50) = 22 \text{ dollars per unit} \]
Verify: notice which x the derivative was evaluated at
Why: It was evaluated at 50, not 51: the 51st unit is the one taking production FROM 50 TO 51, so the rate that estimates it is the rate at the start of that step. Evaluating at 51 would give 22.20, estimating the 52nd unit instead. This off-by-one is the standard error in marginal problems, and reading the phrase 'the nth unit' as 'the step from n minus 1 to n' prevents it.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 232-233
Trap
\[ C'(x) = 12 + 0.2x, \quad \text{cost of the 51st unit} \]
Evaluate at x = 51
Why: The student matches the number in the question.
\[ C'(51) = 22.20 \quad \text{(estimates the 52nd unit)} \]
The 51st unit is produced as output rises from 50 to 51, so the rate at the START of that step is the one that estimates it.
\[ C'(50) = 22 \text{ dollars, estimating } C(51) - C(50) = 22.10 \]
Read 'the nth unit' as the step from n-1 to n
Why: The estimate uses the rate at the beginning of the interval.
The two answers differ by only twenty cents here, which is exactly what makes the error easy to miss and easy to live with. In a steeply curving cost function the discrepancy can be substantial, and the habit of naming the interval before evaluating anything is what keeps it right.
Prediction
Commit before reasoning.
Predict first
For a curve bending upward, is the tangent-line estimate too high or too low?
Correct: Too low — the tangent lies beneath a curve that bends upward.
\[ f'' > 0 \;\Longrightarrow\; \text{the curve lies above its tangents} \]
Why: A curve bending upward lies above every one of its tangent lines, so the tangent estimate undershoots on both sides of the point. That the direction is the same on both sides is worth noticing: it depends on the bending, not on which way you step. Chapter 4 will call upward bending concave up and prove this, and it is why the second derivative's sign is what determines whether a linear estimate over- or under-shoots.
Matching
Which input value does the derivative get?
Match the pairs
Why: In every row the derivative is evaluated at the START of the interval and multiplied by the interval's length. The first two show the off-by-one that catches people: the nth unit is the step from n minus 1 to n.
Sorting
The error grows roughly with the square of the step.
Sort into buckets
Sort each situation by how reliable the tangent estimate is.
The linear case is worth dwelling on: the estimate is exact precisely because there is no bending for the tangent to fail to follow. Everything else is a matter of degree, and the degree is set by the second derivative — which is the subject of Section 4.2.
Section
Section 2
Concept
For an object moving on a line, the derivative of position is velocity and the derivative of velocity is acceleration. Speed is the absolute value of velocity: velocity carries a sign recording direction, and speed does not.
velocity, speed and acceleration — Velocity is the derivative of position and carries a sign giving the direction of travel. Speed is its absolute value. Acceleration is the derivative of velocity, or the second derivative of position.
\[ v(t) = s'(t), \qquad \text{speed} = |v(t)|, \qquad a(t) = v'(t) = s''(t) \]
The object is momentarily at rest exactly where the velocity is zero, and it changes direction where the velocity changes sign. Those are different conditions, as a moment's thought about a ball thrown upward shows.
Figure (svg): Position, velocity and speed for a particle that reverses direction
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 233-236 — motion along a line
Picture it
Position, velocity and speed for a particle that reverses.
Figure (svg): Position, velocity and speed for a particle that reverses direction
The middle stretch has negative velocity, which the speed graph reflects above the axis. Reading the sign of the velocity is how you know which way the object is going, and the position graph confirms it by falling there.
Worked example
Example 3.29. Differentiate twice, then read the signs.
\[ \text{A particle has } s(t) = t^3 - 6t^2 + 9t. \text{ Find when it is at rest and when it moves backwards.} \]
Differentiate for the velocity
Why: Termwise.
\[ v(t) = 3 t ^{2} - 12 t + 9 \]
Factor
Why: Take out 3, then factor the quadratic.
\[ 3(t - 1) (t - 3) \]
Set the velocity to zero for rest
Why: The product vanishes.
\[ \text{at rest at } t = 1\text{ and } t = 3 \]
Read the sign between the roots
Why: A upward parabola is negative between its roots.
\[ v < 0\text{ on } (1, 3) \]
Interpret
Why: Negative velocity means backwards.
\[ \text{moves backwards between } 1\text{ and } 3 \]
Figure (svg): The solution to Worked example analysing a motion shown as a ladder of expressions, one row per legal move
\[ v(t) = 3(t-1)(t-3): \; \text{rest at } t = 1, 3; \; \text{backwards on } (1,3) \]
Verify: check the position at the turning times
Why: At t equal to 1 the position is 1 minus 6 plus 9, which is 4; at t equal to 3 it is 27 minus 54 plus 27, which is 0. So the particle advances to 4, reverses back to 0, then advances again — exactly what a velocity negative only between 1 and 3 requires. Note that being at rest at t equal to 1 does NOT mean the particle stopped permanently; it is an instant of zero velocity as the direction reverses, like a ball at the top of its flight.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 234-235
Sorting
Read the sign of the velocity.
Sort into buckets
Sort each condition, for a particle with v(t) = 3(t-1)(t-3).
The two rest instants are not the particle stopping: they are the moments it turns around. A ball at the top of its flight has zero velocity and is very much still in motion, which is the same situation.
Worked example
Checkpoint 3.29. The reversal makes these differ.
\[ \text{For the same particle on } [0, 4], \text{ find the displacement and the total distance travelled.} \]
Compute the displacement
Why: Final position minus initial.
\[ s(4) - s(0) = 4 - 0 = 4 \]
Identify the reversal points
Why: Where the velocity changes sign.
\[ t = 1\text{ and } t = 3 \]
Compute the distance on each stretch
Why: Absolute changes in position.
\[ | 4 - 0 | + | 0 - 4 | + | 4 - 0 | \]
Add them
Why: Four plus 4 plus 4.
\[ 12 \]
Figure (svg): The solution to Worked example distance travelled against displacement shown as a ladder of expressions, one row per legal move
\[ \text{displacement } 4, \qquad \text{distance } 12 \]
Verify: confirm why the two differ so much
Why: The particle goes forward 4, back 4, then forward 4 again — so it covers 12 units of ground while ending only 4 units from where it started. Displacement is a net change and ignores the retracing; distance counts every unit travelled. Splitting at the reversal points is essential: computing the absolute change from 0 to 4 directly would have given 4 and missed the entire middle excursion.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 235-236
Error analysis
A student computes how far a reversing particle travels.
Annotate
On: \( \text{distance} = |s(4) - s(0)| = |4 - 0| = 4 \)
Whenever the velocity changes sign inside the interval, the endpoints alone cannot give the distance travelled. Find the zeros of the velocity first, split there, and add the absolute changes.
Fill the middle
The velocity from the worked example, already factored.
Fill in the blanks
3(t-1)(t-3) = 0 \;\Longrightarrow\; t = 1 \text3 t = ___
Why: The velocity vanishes at t equal to 1 and 3, which are the two instants the particle reverses. Between them the velocity is negative and the particle moves backwards.
Two truths and a lie
All three are about motion.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Zero velocity is an instantaneous condition, and it typically marks a reversal rather than a permanent stop — a ball at the top of its flight, or this particle at t equal to 1. The object is at rest AT THAT INSTANT and moving immediately before and after. Reading an instantaneous condition as a lasting state is the standard error here.
Ranking
Finding total distance travelled over an interval.
Put in order
Why: Step c is the one that is skipped: zeros of the velocity outside the interval are irrelevant and must be discarded, or the interval is split in the wrong places. Skipping steps b through d entirely and using the endpoints gives the displacement, which is a different quantity.
Section
Section 3
Concept
An object speeds up when its velocity and acceleration have the same sign, and slows down when they have opposite signs. A negative acceleration does not by itself mean slowing down.
speeding up and slowing down — Speed increases when velocity and acceleration share a sign, because the push is in the direction of travel. Speed decreases when they differ, because the push opposes the motion.
\[ v \cdot a > 0 \;\Longrightarrow\; \text{speeding up}; \qquad v \cdot a < 0 \;\Longrightarrow\; \text{slowing down} \]
The compact form is the product's sign. Since speed is the absolute value of velocity, what matters is whether the acceleration is pushing the velocity further from zero or back toward it.
Figure (svg): The rule for speeding up and slowing down, as a sign comparison between velocity and acceleration
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 236-238 — speeding up and slowing down
Picture it
The rule, stated as a comparison.
Figure (svg): The rule for speeding up and slowing down, as a sign comparison between velocity and acceleration
The two speeding-up rows have matching signs and the two slowing-down rows have opposite signs. Neither verdict can be read from the acceleration alone, which is what makes the naive reading wrong.
Worked example
Example 3.31. Two signs, then compare.
\[ \text{For } s(t) = t^3 - 6t^2 + 9t, \text{ is the particle speeding up at } t = 2? \]
Find the velocity there
Why: From v(t) = 3t^2 - 12t + 9.
\[ v(2) = 12 - 24 + 9 = -3 \]
Find the acceleration there
Why: Differentiate again: a(t) = 6t - 12.
\[ a(2) = 0 \]
Compare the signs
Why: The acceleration is zero.
Interpret
Why: The speed is momentarily neither growing nor shrinking.
Check a nearby instant
Why: At t equal to 2.5 the velocity is negative and the acceleration positive.
Figure (svg): The solution to Worked example deciding at an instant shown as a ladder of expressions, one row per legal move
\[ v(2) = -3, \; a(2) = 0 \]
Verify: examine the behaviour on both sides
Why: Just before t equal to 2 the acceleration is negative and the velocity is negative, so the signs match and the particle is speeding up. Just after, the acceleration turns positive while the velocity is still negative, so it begins slowing down. So t equal to 2 is where the speed is greatest during the backward excursion — which the speed graph confirms with a peak there. Zero acceleration marks a turning point of SPEED, not of position.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 237-237
Sorting
Compare the two signs.
Sort into buckets
Sort each pair of readings.
Two of the three speeding-up cases have NEGATIVE acceleration, which is exactly the point. The verdict cannot be read off the acceleration alone, and the product's sign is the compact test.
Worked example
Checkpoint 3.31. The case that defeats the naive reading.
\[ \text{At } t = 0.5 \text{ the same particle has } v = -3.75 \text{... check: } v(0.5) = 3.75. \text{ Instead take } t = 2.5. \]
Compute the velocity
Why: Three times 6.25 minus 30 plus 9.
\[ v(2.5) = -2.25 \]
Compute the acceleration
Why: Fifteen minus 12.
\[ a(2.5) = 3 \]
Compare the signs
Why: Negative velocity, positive acceleration.
Conclude
Why: Opposite signs mean the push opposes the motion.
Contrast with t = 1.5
Why: There v is negative and a is negative 3.
Figure (svg): The solution to Worked example negative acceleration, speeding up shown as a ladder of expressions, one row per legal move
\[ v(2.5) < 0, \; a(2.5) > 0 \;\Longrightarrow\; \text{slowing down} \]
Verify: confirm the counterintuitive case
Why: At t equal to 1.5 the acceleration is negative 3 and the particle is nevertheless SPEEDING UP, because it is travelling backwards and being pushed further backwards. Its speed rises from 0 toward its peak at t equal to 2. This is the case the naive reading gets wrong: negative acceleration means slowing down only for an object moving forwards. The rule is about agreement of signs, and it has to be, because speed is an absolute value.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 238-238
Trap
\[ a(1.5) = -3 < 0 \]
Conclude the particle is slowing down
Why: The student reads the acceleration's sign alone.
\[ \text{so the speed is decreasing} \quad \text{(wrong here)} \]
The velocity at that instant is also negative, so the push is in the direction of travel and the speed is rising.
\[ v(1.5) < 0 \text{ and } a(1.5) < 0 \;\Longrightarrow\; \text{same signs} \;\Longrightarrow\; \text{speeding up} \]
Compare the two signs, never one alone
Why: Speed is the absolute value of velocity, so what matters is whether the acceleration pushes it away from zero or toward it.
The words are misleading and the mathematics is not: 'deceleration' is not a synonym for negative acceleration. A car reversing and accelerating backwards has negative velocity and negative acceleration, and it is unambiguously speeding up.
Fill the middle
A particle with negative velocity and negative acceleration.
Fill in the blanks
v = -5, \; a = -2 \;\Longrightarrow\; v \cdot a = 10 > 0 \;\Longrightarrow\; \text___
Why: The product is positive 10, so the signs agree and the particle is speeding up despite the acceleration being negative. The product's sign is the whole test, in one number.
Two truths and a lie
All three are about acceleration.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and the word 'deceleration' is what makes it tempting. Negative acceleration means slowing down only for forward motion. When the velocity is also negative the two agree and the object speeds up backwards, which is what a reversing car accelerating does. The rule must compare signs, because speed is an absolute value and cannot see direction.
Prediction
Commit before reasoning.
Predict first
Why does the rule compare the signs of velocity and acceleration rather than reading the acceleration alone?
Correct: Because speed is an absolute value, so the question is whether the push moves the velocity away from zero.
\[ \frac{d}{dt}|v| = \frac{v}{|v|}\cdot a, \quad \text{which is positive exactly when } va > 0 \]
Why: Speed cannot see direction, so it grows exactly when the velocity's magnitude grows — which happens when the acceleration pushes it further from zero, that is, in the same direction it already points. Whether that direction is positive or negative is irrelevant to speed. This is why the rule is a comparison, and it is also why the compact form is the sign of the product: a positive product is precisely two agreeing signs.
Section
Section 4
Concept
Marginal cost is the derivative of the cost function, marginal revenue of revenue, and marginal profit of profit. Each estimates the effect of producing and selling one additional unit.
marginal cost, revenue and profit — The derivatives of the cost, revenue and profit functions with respect to the quantity produced. Each approximates the change caused by one more unit, and each has units of currency per unit.
\[ MC(x) = C'(x), \quad MR(x) = R'(x), \quad MP(x) = P'(x) = R'(x) - C'(x) \]
Because profit is revenue minus cost, marginal profit is marginal revenue minus marginal cost. Profit stops rising exactly where those two are equal, which is the optimisation condition Section 4.7 will exploit.
Figure (svg): Cost, revenue and profit curves with their marginal quantities as tangent slopes
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 238-240 — marginal functions in economics
Picture it
Cost rising faster than linearly, against linear revenue.
Figure (svg): Cost, revenue and profit curves with their marginal quantities as tangent slopes
Revenue here is a straight line, so marginal revenue is constant. Cost curves upward, so marginal cost rises — and where it overtakes marginal revenue, each further unit costs more than it earns.
Worked example
Example 3.33. Differentiate each, then compare.
\[ \text{With } C(x) = 1000 + 12x + 0.1x^2 \text{ and } R(x) = 40x, \text{ find the marginal profit at } x = 100. \]
Differentiate the cost
Why: Termwise.
\[ C'(x) = 12 + 0.2 x \]
Differentiate the revenue
Why: A constant slope.
\[ R'(x) = 40 \]
Form the marginal profit
Why: Revenue's rate minus cost's rate.
\[ P'(x) = 40 - (12 + 0.2 x) \]
Simplify
Why: Collect.
\[ P'(x) = 28 - 0.2 x \]
Evaluate
Why: At 100 units.
\[ P'(100) = 8\text{ dollars per unit} \]
Figure (svg): The solution to Worked example marginal cost and revenue shown as a ladder of expressions, one row per legal move
\[ P'(100) = 8 \text{ dollars per unit} \]
Verify: find where the marginal profit vanishes
Why: Setting 28 minus 0.2x to zero gives x equal to 140, so profit is still rising at 100 and stops rising at 140. Beyond that each further unit loses money, because marginal cost has overtaken the constant marginal revenue of 40. Checking against the exact difference: P(101) minus P(100) is 7.90, close to the predicted 8. The whole analysis rests on the fact that marginal profit is positive up to 140 and negative after — which is exactly the optimisation of Section 4.7.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 239-240
Matching
Each estimates one additional unit.
Match the pairs
Why: The last row is where this section points: setting the marginal profit to zero locates the production level that maximises profit, which is equivalent to marginal revenue equalling marginal cost. That equivalence is one of the standard results of economics and it is pure calculus.
Worked example
Checkpoint 3.33. What a negative marginal profit means.
\[ \text{Interpret } P'(200) \text{ for the same firm.} \]
Evaluate the marginal profit
Why: Twenty-eight minus 40.
\[ P'(200) = -12 \]
Read the sign
Why: Negative.
Attach the units
Why: Dollars per unit.
\[ -12\text{ dollars per unit} \]
Interpret for the firm
Why: Producing the 201st unit reduces profit.
Figure (svg): The solution to Worked example interpreting the sign shown as a ladder of expressions, one row per legal move
\[ P'(200) = -12 \text{ dollars per unit} \]
Verify: distinguish falling profit from a loss
Why: Negative marginal profit does NOT mean the firm is losing money — total profit at 200 units is R minus C, which is 8000 minus 7400, a profit of 600 dollars. What is negative is the RATE: the firm is past its best output and each additional unit erodes that 600. Confusing a negative derivative with a negative value is the standard error, and it is the same one-rung-down mistake as reading falling inflation as falling prices.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 240-240
Error analysis
A student interprets a firm's position at 200 units.
Annotate
On: \( P'(200) = -12 \;\Longrightarrow\; \text{the firm is losing } 12 \text{ dollars} \)
The units settle it: dollars per unit is a rate and dollars is an amount. A negative rate on a positive quantity means the quantity is shrinking from a healthy level, which is a completely different situation from a loss.
Fill the middle
The firm from the worked example, with both rates computed.
Fill in the blanks
P'(x) = 40 - (12 + 0.2x) = 28 - 0.2x
Why: Marginal profit is marginal revenue minus marginal cost, giving 28 minus 0.2x. It vanishes at x equal to 140, which is the profit-maximising output.
Prediction
Commit before reasoning.
Predict first
A firm finds P'(140) = 0. What has it found?
Correct: The output where profit stops rising: marginal revenue equals marginal cost there.
\[ P'(x) = 0 \iff R'(x) = C'(x) \quad \text{- the classic optimality condition} \]
Why: A zero derivative marks a horizontal tangent on the profit curve, which here is its peak. Total profit at that output is substantial, not zero — the derivative being zero says the profit is momentarily not changing, which is precisely what a maximum looks like. The break-even point is where profit itself is zero, an entirely different condition found by solving P equals zero. This is the same distinction between a value and a rate that the previous idea's error turned on.
Sorting
Check the units.
Sort into buckets
Sort each quantity for a firm.
The pair c and d describe the same firm at the same output and say opposite-sounding things: profit is 600 dollars and falling at 12 dollars per unit. Both are true, and keeping the value and the rate apart is what makes them consistent.
Section
Section 5
Concept
A population's growth rate, a reaction's rate, a tank's filling rate and a bank balance's growth are all derivatives. The mathematics is identical and only the units change — and the units carry the interpretation.
units of a derivative — Always the output's units divided by the input's. Differentiating twice with respect to the same variable squares the denominator's unit, which is why acceleration is measured per second squared.
\[ \left[\frac{dy}{dx}\right] = \frac{[y]}{[x]} \]
Units are the cheapest error check available. An answer whose units are wrong is wrong before any arithmetic is examined, and an answer's units often reveal which quantity was actually computed.
Figure (svg): A table of quantities and the units their derivatives carry
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 232-240 — rates of change in the sciences
Picture it
Four quantities and the units their derivatives carry.
Figure (svg): A table of quantities and the units their derivatives carry
The second row is the informative one: differentiating a velocity with respect to time divides by seconds again, producing the squared unit. That is where metres per second squared comes from, and it is not a convention.
Worked example
Example 3.34. Differentiate, evaluate, and attach units.
\[ \text{A population is } P(t) = 400 + 30t^2 - t^3 \text{ after } t \text{ years. Find } P'(4) \text{ and interpret.} \]
Differentiate
Why: Termwise.
\[ P'(t) = 60 t - 3 t ^{2} \]
Evaluate at the given time
Why: 240 minus 48.
\[ P'(4) = 192 \]
Attach the units
Why: Individuals per year.
\[ 192\text{ per year} \]
Interpret
Why: The sign is positive.
\[ \text{growing by about } 192\text{ in the next year} \]
Figure (svg): The solution to Worked example a population growth rate shown as a ladder of expressions, one row per legal move
\[ P'(4) = 192 \text{ per year} \]
Verify: check against the actual next year
Why: The population at t equal to 4 is 400 plus 480 minus 64, which is 816; at t equal to 5 it is 400 plus 750 minus 125, which is 1025. The actual increase is 209, against the predicted 192 — reasonably close, and the estimate is low because the population is still accelerating over that year. Note also that P prime vanishes at t equal to 20, so this model predicts the population peaks there and declines afterwards, which is worth knowing before extrapolating it.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 240-240
Matching
Output units over input units.
Match the pairs
Why: The second row shows the squared unit arising naturally: dividing by seconds twice. Nothing about metres per second squared is conventional — it is what the definition of a second derivative produces.
Worked example
Checkpoint 3.34. The units are a complete check.
\[ \text{A tank's volume is } V(t) \text{ litres after } t \text{ minutes. A student reports } V'(3) = 40 \text{ litres. Is that right?} \]
Identify the output and input units
Why: Litres and minutes.
Deduce the derivative's units
Why: Output over input.
Compare with the reported units
Why: The student wrote litres.
Diagnose
Why: A volume has been reported where a rate was asked for.
\[ \text{likely } V(3),\text{ not } V'(3) \]
Figure (svg): The solution to Worked example using units to catch an error shown as a ladder of expressions, one row per legal move
\[ \left[V'\right] = \frac{\text{L}}{\text{min}}, \text{ not L} \]
Verify: consider what the mismatch usually indicates
Why: Reporting the output's own units almost always means the function was evaluated instead of its derivative — the student computed V of 3 rather than V prime of 3. That diagnosis comes free from the units, without seeing any of the work. This is why attaching units to every applied answer is worth the few seconds: it catches not just arithmetic slips but whole-question misreadings.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 233-233
Trap
\[ P'(4) = 192 \]
Report the bare number
Why: The student gives a value with no units attached.
Is that 192 individuals, 192 per year, or 192 percent? The number alone does not say, and the three readings describe very different situations.
\[ P'(4) = 192 \text{ individuals per year} \]
Attach output units over input units, always
Why: The units are part of the answer, not decoration.
In an applied problem the units are frequently the whole content: 192 individuals per year and 192 individuals are different claims, and only one of them answers a question about a rate. They are also the fastest check available — a mismatched unit reveals the wrong quantity was computed before any arithmetic is examined.
Fill the middle
The population model from the worked example, differentiated.
Fill in the blanks
P'(t) = 60t - 3t^2 \;\Longrightarrow\; P'(4) = 240 - 48 = 192
Why: Three times 4 squared is 48, so the rate is 192 individuals per year. The population is growing, and the units are what make that a rate rather than a headcount.
Two truths and a lie
All three are about units.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. A position in metres has a derivative in metres per second — a genuinely different kind of quantity. Reporting a derivative in the function's own units is the standard sign that the function was evaluated instead of its derivative, which is exactly what the tank example diagnosed.
Prediction
Commit before reasoning.
Predict first
A population model gives P'(25) = -75 individuals per year. What follows?
Correct: The population is shrinking by about 75 per year, and may still be large.
\[ P(25) = 3525 > 0 \text{ while } P'(25) = -75 < 0 \]
Why: A negative rate says the quantity is decreasing, not that the quantity is negative — those are the value and the rate again, one rung apart. For this model the population at t equal to 25 is 400 plus 18750 minus 15625, which is 3525 individuals, and it is falling. Whether the model has failed is a separate judgement: a declining population is perfectly realistic, though a model predicting the population reaching zero should be treated with suspicion beyond that point.
Comparison
Fill the blanks. One piece of mathematics, four vocabularies.
Comparison matrix
| Field | The derivative is called | Its units |
|---|---|---|
| Physics | velocity | metres per second |
| Physics, twice | acceleration | metres per second squared |
| Economics | marginal cost | dollars per unit |
| Biology | growth rate | individuals per year |
Nothing about the mathematics changes between rows. What changes is the vocabulary and the units, and the units are what let you check that the right quantity was computed.
Pattern
Given an applied problem asking about a rate.
Steps one and five bracket the whole procedure and are both about units. Together they catch the two commonest failures: computing the value instead of the rate, and evaluating at the wrong end of an interval.
Stewart, Calculus: Early Transcendentals 8e, §3.7 Rates of Change in the Natural and Social Sciences §3.7, pp. 224-236
Check
Marginal quantities. Evaluate at the start of the step.
Check your understanding
With C(x) = 1000 + 12x + 0.1x^2, estimate the cost of the 51st unit.
Answer: A
Why: C'(x) = 12 + 0.2x, and the 51st unit takes production from 50 to 51, so evaluate at 50.
Check
Speeding up. Compare the two signs.
Check your understanding
A particle has v = -5 and a = -2. What is it doing?
Answer: A
Why: Both signs are negative, so they agree: the push is in the direction of travel and the speed grows.
Check
Units. Output over input.
Check your understanding
A tank's volume V is in litres and time t in minutes. What are the units of V'(t)?
Answer: A
Why: A derivative always carries the output's units divided by the input's.
Real world
A hospital tracks a patient's blood oxygen saturation, in percent, minute by minute. At one reading the saturation is 91 percent and the rate of change is negative 0.4 percent per minute; ten minutes later the saturation is 88 percent and the rate is negative 0.1 percent per minute.
Discussion prompt
Which reading is more concerning, and why? Estimate how long until saturation reaches 85 percent under each rate, and say what the change in the rate itself indicates.
Hint: Compare the values, the rates, and how the rates are changing.
Answer:
The values say the second reading is worse: 88 is lower than 91, and further below the safe threshold.
The rates say the opposite about the trend. At negative 0.4 per minute the fall is four times faster than at negative 0.1.
\[ \text{from } 91: \; \frac{91-85}{0.4} = 15 \text{ min}; \qquad \text{from } 88: \; \frac{88-85}{0.1} = 30 \text{ min} \]
So despite the lower value, the second reading gives twice as long before the critical threshold — because the deterioration has slowed markedly.
The change in the rate is the second derivative, and it is positive: negative 0.4 rising to negative 0.1. The patient is still declining, but the decline is decelerating, which is exactly what a treatment beginning to work looks like.
The clinical judgement needs all three rungs. The value says how bad things are now, the rate says how fast they are getting worse, and the change in the rate says whether the intervention is working. Reading only one rung — which is what 'saturation is 88' does — discards most of the information the monitor is providing.
\[ \text{value } 88, \quad \text{rate } -0.1, \quad \text{rate of the rate } > 0 \]
Commit first
Answer, then rate your confidence honestly.
Predict first
An object has negative velocity and negative acceleration. Is it speeding up or slowing down?
Correct: Speeding up — the two signs agree.
\[ v \cdot a = (-5)(-2) = 10 > 0 \;\Longrightarrow\; \text{speeding up} \]
Why: Speed is the absolute value of velocity, so it grows whenever the acceleration pushes the velocity further from zero. With both negative, the object is moving backwards and being pushed further backwards, so its speed rises. A car reversing while accelerating backwards is the everyday case. The word deceleration is what makes the first option tempting, and it is precisely why the rule compares signs rather than reading either one alone. Position plays no part at all.
Explain it
They insist that negative acceleration always means slowing down, because that is what deceleration means.
Discussion prompt
In four sentences or fewer, give them a physical case that settles it.
Hint: Put them in a car that is reversing.
Answer:
Ask them to imagine reversing out of a driveway and pressing the accelerator harder. The car is moving backwards, so its velocity is negative, and it is gaining speed backwards, so its acceleration is negative too. Both are negative and the car is unmistakably speeding up.
The trouble is the word: 'deceleration' means the speed is dropping, which is not the same as the acceleration being negative. Speed cannot see direction, so what matters is whether the push agrees with the motion — same signs speed you up, opposite signs slow you down, whichever way you happen to be pointing.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For marginal quantities, read 'the nth unit' as the step from n minus 1 to n. For distance, find the velocity's zeros first and split there. For speeding up, compute the product of velocity and acceleration and read its sign. For units, write output over input before evaluating anything, and check the answer against the kind of quantity asked for. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a curve with its tangent at a point, step one unit along, and mark the tangent estimate, the true value and the gap between them; write beneath it why the estimate is low for an upward-bending curve. Below, take the position function t cubed minus 6t squared plus 9t and draw three stacked graphs with a common time axis: position, velocity and speed. Mark the two instants of rest, shade the stretch where the motion is backwards, and write the displacement and the total distance on the interval from 0 to 4 with the split points shown. Beside that, write the four sign combinations of velocity and acceleration in a two-by-two table, marking which two mean speeding up. In the lower half, take the cost function 1000 plus 12x plus a tenth of x squared and the revenue 40x; write the marginal cost, marginal revenue and marginal profit, find where marginal profit vanishes, and write one sentence saying what that output means. In a margin, write the units of four different derivatives, each as output over input.
If your total distance equals your displacement, you have not split at the velocity's zeros — this particle covers 12 units of ground while finishing only 4 from where it started, and the difference is the entire point of the exercise.
Recap
Five things, and none of them is a new rule. They are what the rules were for.
| If you see | Then |
|---|---|
| 'The nth unit' | Evaluate the derivative at n - 1 |
| A reversal inside the interval | Split there for total distance |
| Zero velocity | An instant of rest, usually a reversal |
| Velocity and acceleration agreeing in sign | Speeding up |
| 'Marginal' anything | It is a derivative, in currency per unit |
| A negative rate | The quantity is falling, not negative |
| An answer without units | It is not yet an answer |
Section 3.5 returns to rule-building with the trigonometric functions. Their derivatives depend on the limit of sine over x that Section 2.3 established by squeezing, which is where the insistence on radians finally pays.
OpenStax Calculus Volume 1, §3.4 Derivatives as Rates of Change §3.4, pp. 230-240 — everything on these slides traces back here
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