The Meaning of the Derivative

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$f(a+h)\approx f(a)+f'(a)\,h$

include the units in the sentence

What to Remember

Say the rate, the moment, and the units

If $W$ is gallons and $t$ is hours, then $W'(3) = -4$ means this: at $t = 3$ hours, the amount of water is decreasing at a rate of 4 gallons per hour.

The sign is the direction of change

A positive derivative means the quantity is increasing at that instant. A negative derivative means it is decreasing. The number you report in the sentence is the absolute value, and the word increasing or decreasing carries the sign.

A derivative estimates a small change

$f(a + h) \approx f(a) + f'(a)\cdot h$

The estimate uses the rate at $a$ as if it stayed constant over that step.

Average rate and instantaneous rate answer different questions

The average rate on $[a, b]$ is $\dfrac{f(b) - f(a)}{b - a}$. The instantaneous rate is $f'(a)$.

Level 1

Easy

Translate the derivative into a sentence with units.

Example

The temperature of a cup of tea, in degrees Fahrenheit, is $T(t)$, where $t$ is the number of minutes after the tea is poured. $T'(4) = -3$. What does that mean?

$T$ is temperature in °F, and $t$ is minutes.

$T'(4)$ is a rate in °F per minute at $t = 4$.

The value is negative, so the temperature is decreasing.

At 4 minutes, the temperature is decreasing at a rate of 3°F per minute.

Answer: decreasing at 3°F per minute at $t = 4$

The number of people in a park is $P(t)$, where $t$ is hours after noon. $P'(2) = 40$. What does that mean?

Show Answer

Units of $P'$ are people per hour.

The value is positive, so the number of people is increasing.

At 2 hours after noon, the number of people is increasing at 40 people per hour.

Answer: increasing at 40 people per hour

$C(x)$ is the cost, in dollars, of producing $x$ items. $C'(50) = 2.4$. What does that mean?

Show Answer

Units of $C'$ are dollars per item.

When 50 items have been produced, the cost is increasing at $2.40 per additional item.

Answer: $2.40 per item at 50 items

$V(t)$ is the volume of water in a tank, in liters, $t$ minutes after a drain is opened. $V'(10) = -6$. What does that mean?

Show Answer

The rate is negative.

At $t = 10$ minutes, the volume is decreasing at 6 liters per minute.

Answer: decreasing at 6 L/min

$A(t)$ is the area of an algae patch, in square meters, after $t$ days. What are the units of $A'(t)$?

A. square meters per day

B. square meters

C. days

D. meters per day

Show Answer

$A$ is square meters and $t$ is days.

A derivative with respect to time has units of output per input.

Answer: A. square meters per day

At $x = 5$, $f'(5) < 0$. What is happening to $f$ at that input?

Show Answer

A negative derivative means the function is decreasing at that input.

Answer: f is decreasing at $x = 5$

Level 2

Medium

Use the rate at one moment to estimate a nearby value.

Example

Revenue is measured in dollars, and $t$ is the day number. $R'(8) = 15$. Estimate the change in revenue from day 8 to day 8.5.

$\Delta R \approx R'(8)\cdot\Delta t$

$\Delta t = 0.5$

$15\cdot 0.5$

$7.5$

Revenue increases by about 7.5 dollars.

Answer: about 7.5 dollars

$W(5) = 200$ gallons and $W'(5) = -4$ gallons per hour. Estimate $W(5.25)$.

Show Answer

$W(5.25) \approx W(5) + W'(5)\cdot 0.25$

$200 + (-4)(0.25)$

$200 - 1$

$199$

Answer: about 199 gallons

A car's position, in meters, has $s'(3) = 12$ meters per second. Estimate the change in position from $t = 3$ to $t = 3.1$.

Show Answer

$\Delta s \approx s'(3)\cdot 0.1$

$12\cdot 0.1$

$1.2$

Answer: about 1.2 meters

The depth of snow, in meters, is $H(t)$ after $t$ hours. $H'(6) = -0.2$. Which statement matches that value?

A. At 6 hours, the depth is decreasing at 0.2 meters per hour.

B. At 6 hours, the depth is 0.2 meters.

C. At 6 hours, the depth is decreasing at 0.2 meters.

D. The depth decreased by 0.2 meters during the first 6 hours.

Show Answer

$H'(6)$ is a rate at one moment, in meters per hour.

The negative sign means the depth is decreasing.

Answer: A. At 6 hours, the depth is decreasing at 0.2 meters per hour.

A population is measured in thousands of people, and $t$ is years. $P'(10) = -1.5$. Describe the population at $t = 10$.

Show Answer

One unit of $P$ is 1 thousand people.

$-1.5$ thousand people per year is 1500 people per year.

At $t = 10$, the population is decreasing at 1500 people per year.

Answer: decreasing by 1500 people per year

Use the table. $V$ is gallons of water in a pool and $t$ is minutes. Find the average rate of change of $V$ from $t = 0$ to $t = 4$, and include units.

$t$ (min)	0	4	8	12
V (gal)	10	18	22	23

Show Answer

$\dfrac{V(4) - V(0)}{4 - 0}$

$\dfrac{18 - 10}{4}$

$\dfrac{8}{4}$

$2$ gallons per minute

Answer: 2 gallons per minute

Level 3

Hard

An FRQ-style reading: interpret the rate, estimate a value, and compare it with an average.

Example

$E(t)$ is the energy stored in a battery, in joules, $t$ minutes after charging begins. $E(12) = 4000$ and $E'(12) = 30$. Interpret $E'(12)$, then estimate $E(12.4)$.

At 12 minutes, the stored energy is increasing at 30 joules per minute.

$E(12.4) \approx E(12) + E'(12)\cdot 0.4$

$4000 + 30\cdot 0.4$

$4000 + 12$

$4012$

4012 joules is an estimate that holds the rate at $t = 12$ constant for 0.4 minutes.

Answer: increasing at 30 J/min; about 4012 joules

$W(10) = 40$ gallons and $W'(10) = -1.5$ gallons per minute. Estimate $W(12)$.

Show Answer

$W(12) \approx 40 + (-1.5)(2)$

$40 - 3$

$37$

This is an estimate. It uses the rate at $t = 10$ for the whole two minutes.

Answer: about 37 gallons

Use the table of $V$ in gallons. Find the average rate of change from $t = 8$ to $t = 12$.

$t$ (min)	0	4	8	12
V (gal)	10	18	22	23

Show Answer

$\dfrac{23 - 22}{12 - 8}$

$\dfrac{1}{4}$

0.25 gallons per minute

Answer: 0.25 gallons per minute

Use the same table. The average rate from $t = 0$ to $t = 4$ is 2 gallons per minute, and the average rate from $t = 8$ to $t = 12$ is 0.25 gallons per minute. $V$ is still increasing. What do those two average rates say?

$t$ (min)	0	4	8	12
V (gal)	10	18	22	23

Show Answer

Both average rates are positive, so $V$ increases on each of those intervals.

The later average rate is smaller.

Over these intervals, water is still being added, and the average rate of change is smaller on the later interval.

Answer: still increasing, at a smaller average rate

A tank contains $W(t)$ gallons after $t$ minutes. $W(5) = 20$ and $W'(5) = 2$. Which statement must be true?

A. At $t = 5$, the volume is increasing at 2 gallons per minute.

B. At $t = 5$, the tank contains 2 gallons.

C. The tank gained 2 gallons during the first 5 minutes.

D. The volume at $t = 6$ is exactly 22 gallons.

Show Answer

$W(5) = 20$ is the amount at that minute.

$W'(5) = 2$ is the rate at that minute.

The rate gives an estimate for a later time. It does not lock the value at $t = 6$.

Answer: A. increasing at 2 gal/min at $t = 5$

Cost $C$ is in dollars and $n$ is the number of shirts. $C(40) = 600$ and $C'(40) = 8$. Estimate the cost of producing 42 shirts, and state the units of $C'(40)$.

Show Answer

Units of $C'(40)$ are dollars per shirt.

$C(42) \approx 600 + 8\cdot 2$

$616$

About $616.

Answer: dollars per shirt; about $616

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