Derivatives in Context

A derivative is a rate of change: it tells you how fast one quantity is changing compared with another. On the AP exam you will often be asked to explain what a number like $V'(5) = -12$ means in a real situation, with the right units. This page shows you how to write those sentences clearly, and how to find and estimate rates in contexts like temperature, population, volume, and cost.

Key ideas

What f’(a) means

If $y = f(x)$, then $f'(a)$ is the instantaneous rate of change of $y$ with respect to $x$ when $x = a$. It’s the slope of the tangent line at $x = a$.

• If $f'(a) \gt 0$, the quantity is increasing at that moment.
• If $f'(a) \lt 0$, the quantity is decreasing at that moment.
• The size $|f'(a)|$ tells you how fast.

Units

The units of $f'(x)$ are always

$$\frac{\text{units of the output}}{\text{units of the input}}$$

So if $V(t)$ is a volume in litres and $t$ is in minutes, $V'(t)$ is in litres per minute. The second derivative $V''(t)$ is in litres per minute per minute (L/min²): it tells you how fast the rate itself is changing.

The AP-style sentence

A complete interpretation answers four things: what is changing, when (or where), how fast (with units), and which direction (increasing or decreasing). A reliable template:

At $t = a$ (with units), the [quantity] is [increasing/decreasing] at a rate of $|f'(a)|$ [output units] per [input unit].

For example, if $V'(5) = -12$: “At $t = 5$ minutes, the volume of water in the tank is decreasing at a rate of $12$ litres per minute.”

Notice the sentence says “decreasing at a rate of $12$”, not “increasing at a rate of $-12$”. Both are correct, but the first is clearer.

Rates in other contexts

Derivatives describe far more than motion:

Function	Derivative means	Units of the derivative
$P(t)$, population (people), $t$ in years	how fast the population is growing	people per year
$H(t)$, temperature (°C), $t$ in minutes	how fast it’s heating or cooling	°C per minute
$C(x)$, cost (dollars) to make $x$ items	marginal cost: approx. cost of one more item	dollars per item
$V(r)$, volume (cm³), radius $r$ in cm	how fast volume grows as the radius grows	cm³ per cm

Estimating from a table

If you only have a table of values, estimate $f'(a)$ with the slope between the two data points on either side of $a$ (the closest ones that surround it):

$$f'(a) \approx \frac{f(x_2) - f(x_1)}{x_2 - x_1}, \qquad x_1 \lt a \lt x_2$$

This is an average rate of change used as an estimate of the instantaneous rate. Always include units.

Worked examples

Example 1: Writing the sentence

Water drains from a tank. $V(t)$ is the volume of water in litres, $t$ minutes after draining starts. Interpret $V'(5) = -12$.

Solution. The derivative is negative, so the volume is decreasing. Units: litres per minute.

At $t = 5$ minutes, the volume of water in the tank is decreasing at a rate of $12$ litres per minute.

Example 2: Cooling coffee

The temperature of a cup of coffee, in °C, is $H(t) = 20 + 70e^{-0.05t}$, where $t$ is in minutes. Find $H'(10)$ and interpret it.

Solution. Use the chain rule on the exponential:

$$H'(t) = 70e^{-0.05t} \cdot (-0.05) = -3.5e^{-0.05t}$$ $$H'(10) = -3.5e^{-0.5} \approx -2.123$$

At $t = 10$ minutes, the temperature of the coffee is decreasing at a rate of about $2.123$ °C per minute.

$H'(10)$ is the slope of the tangent line at $t = 10$: about $-2.123$ °C per minute.

Example 3: Estimating from a table

A town’s population $P(t)$, in thousands, is recorded $t$ years after 2015.

$t$ (years)	$0$	$2$	$5$	$9$
$P(t)$ (thousands)	$12.0$	$13.1$	$14.9$	$17.3$

Estimate $P'(3.5)$ and interpret it.

Solution. The closest data points on either side of $t = 3.5$ are $t = 2$ and $t = 5$:

$$P'(3.5) \approx \frac{P(5) - P(2)}{5 - 2} = \frac{14.9 - 13.1}{3} = 0.6$$

At $t = 3.5$ years (partway through 2018), the population is increasing at a rate of about $0.6$ thousand people per year, or about $600$ people per year.

Example 4: Marginal cost

It costs $C(x) = 4000 + 150x - 0.05x^2$ dollars to build $x$ bicycles. Find $C'(100)$ and interpret it.

Solution.

$$C'(x) = 150 - 0.1x \quad\Rightarrow\quad C'(100) = 150 - 10 = 140$$

When $100$ bicycles have been built, the cost is increasing at a rate of $140 per bicycle. In other words, building the $101$st bicycle costs about $140.

Common mistakes

Leaving out the units, or using the wrong ones. The units are always output units per input unit. If $P$ is in thousands of people and $t$ in years, $P'$ is in thousands of people per year, not “people” or “years”.

Forgetting “at time $t = a$”. AP graders want the moment. “The volume is decreasing at $12$ litres per minute” is incomplete without “at $t = 5$ minutes”.

Describing the quantity instead of its rate. $V'(5) = -12$ does not mean there are $-12$ litres, or that $12$ litres are left. It’s how fast the volume is changing.

Saying “increasing at a rate of $-12$”. Use the sign to choose the word: negative means decreasing. Then give the size as a positive number.

Estimating with points that don’t surround the input. To estimate $P'(3.5)$, use the data on both sides of $3.5$ (here $t = 2$ and $t = 5$), not $t = 0$ and $t = 2$.

Mixing up the units of $f''$. The second derivative has the input unit twice in the denominator, such as litres per minute per minute.

Practice

1. (Warm-up) Give the units of the derivative in each case.

• (a) $d(t)$ is distance travelled in kilometres, $t$ in hours. Units of $d'(t)$?
• (b) $N(p)$ is the number of concert tickets sold when the price is $p$ dollars. Units of $N'(p)$?

Solution

(a) Kilometres per hour.

(b) Tickets per dollar.

2. (Warm-up) $h(t)$ is the height of a sunflower in centimetres, $t$ weeks after planting. Interpret $h'(3) = 8$.

Solution

At $t = 3$ weeks, the height of the sunflower is increasing at a rate of $8$ centimetres per week.

3. (Warm-up) $A(t)$ is the area of an oil slick in square metres, $t$ hours after a spill. Interpret $A'(2) = 35$, and say what $A(2) = 150$ means. How are the two statements different?

Solution

$A'(2) = 35$: at $t = 2$ hours, the area of the oil slick is increasing at a rate of $35$ square metres per hour.

$A(2) = 150$: at $t = 2$ hours, the oil slick covers $150$ square metres.

$A(2)$ is how big the slick is; $A'(2)$ is how fast it’s growing.

4. (Core) The volume of a spherical balloon is $V(r) = \dfrac{4}{3}\pi r^3$ cm³, where $r$ is the radius in cm. Find $V'(5)$ and interpret it.

Solution

$$V'(r) = 4\pi r^2 \quad\Rightarrow\quad V'(5) = 100\pi \approx 314.159$$

When the radius is $5$ cm, the volume is increasing at a rate of $100\pi \approx 314.159$ cm³ per centimetre of radius.

5. (Core) An oven’s temperature $T(t)$, in °C, is measured $t$ minutes after it is turned on.

$t$ (min)	$0$	$4$	$10$	$15$
$T(t)$ (°C)	$22$	$85$	$160$	$190$

Estimate $T'(7)$, with units, and interpret it.

Solution

Use the data on either side of $t = 7$, which are $t = 4$ and $t = 10$:

$$T'(7) \approx \frac{160 - 85}{10 - 4} = \frac{75}{6} = 12.5$$

At $t = 7$ minutes, the oven’s temperature is increasing at a rate of about $12.5$ °C per minute.

6. (Core) The water depth at a dock is $D(t) = 5 + 2\sin\left(\dfrac{\pi t}{6}\right)$ metres, $t$ hours after midnight (the angle is in radians). Find $D'(4)$ and interpret it.

Solution

$$D'(t) = 2\cos\left(\frac{\pi t}{6}\right) \cdot \frac{\pi}{6} = \frac{\pi}{3}\cos\left(\frac{\pi t}{6}\right)$$$$D'(4) = \frac{\pi}{3}\cos\left(\frac{2\pi}{3}\right) = \frac{\pi}{3}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \approx -0.524$$

At 4:00 a.m. ($t = 4$ hours), the water depth is decreasing at a rate of about $0.524$ metres per hour.

7. (Core) $W(t)$ is the amount of water in a reservoir, in megalitres (ML), $t$ days after June 1. Give the units of $W'(t)$ and $W''(t)$, and interpret $W'(6) = 4$ and $W''(6) = -3$ together.

Solution

$W'(t)$ is in ML per day; $W''(t)$ is in ML per day per day (ML/day²).

At $t = 6$ days, the amount of water is increasing at a rate of $4$ ML per day, and that rate is decreasing by $3$ ML per day per day. The reservoir is still filling, but more slowly.

8. (Challenge) A furniture maker’s cost to build $x$ chairs is $C(x) = 2000 + 40x - 0.02x^2$ dollars.

• (a) Find $C'(200)$ and interpret it.
• (b) Find the actual cost of building the $201$st chair, $C(201) - C(200)$, and compare.

Solution

(a) $C'(x) = 40 - 0.04x$, so $C'(200) = 40 - 8 = 32$. When $200$ chairs have been built, the cost is increasing at a rate of $32 per chair.

(b)

$$C(201) - C(200) = 40(1) - 0.02(201^2 - 200^2) = 40 - 0.02(401) = 31.98$$

The $201$st chair actually costs $31.98, very close to the marginal cost of $32. That’s why marginal cost is used as “the cost of one more item”.

9. (Challenge) A bacteria culture has $P(t) = 1500e^{0.04t}$ cells, $t$ hours after it starts. At what time is the population growing at a rate of $100$ cells per hour? Give an exact answer and a decimal to 3 places.

Solution

$$P'(t) = 1500(0.04)e^{0.04t} = 60e^{0.04t}$$

Set it equal to $100$:

$$60e^{0.04t} = 100 \quad\Rightarrow\quad e^{0.04t} = \frac{5}{3} \quad\Rightarrow\quad t = \frac{\ln(5/3)}{0.04} = 25\ln\frac{5}{3} \approx 12.771$$

After about $12.771$ hours, the population is growing at $100$ cells per hour.