A derivative is a rate, and a rate has units AB & BC

Every derivative is a limit of difference quotients, and a difference quotient is one quantity divided by another:

$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}.$$

The numerator carries the units of $f$. The denominator carries the units of the input. Dividing them gives the rule with no memorizing at all:

The units of $f'$ are the units of $f$ per unit of the input.

1. $V(t)$ in gallons, $t$ in minutes $\Rightarrow$ $V'(t)$ in gallons per minute.
2. $C(x)$ in dollars, $x$ in units produced $\Rightarrow$ $C'(x)$ in dollars per unit.
3. $T(h)$ in degrees Celsius, $h$ in hours $\Rightarrow$ $T'(h)$ in degrees Celsius per hour.

Apply the same rule twice for a second derivative. If $P(t)$ is a population in thousands of people and $t$ is in years, then $P'(t)$ is in thousands of people per year, and $P''(t)$ is in thousands of people per year, per year. That doubled unit is not a typo: it measures how fast the growth rate itself is changing.

A full interpretation of $f'(a) = k$ has four parts, and leaving any one out is what costs the point.

1. When. At the input value $a$, stated with its own units. Not "in general" and not "over the interval."
2. What. The quantity $f$ measures, named in words.
3. Which direction. Increasing if $k > 0$, decreasing if $k < 0$.
4. How fast, with units. The size $|k|$, followed by units of $f$ per unit of input.

So $C'(200) = 4.50$, for $C(x)$ the cost in dollars of producing $x$ units, becomes: when 200 units have been produced, the cost is increasing at a rate of 4.50 dollars per unit.

Notice what the sentence does not say. It does not say the cost is 4.50 dollars, and it does not say the 200th unit cost exactly 4.50 dollars to make. It reports a rate at an instant, and it says so out loud.

$f(a)$ and $f'(a)$ answer two questions that sound alike and are not.

1. $f(a)$: how much there is at $a$. Units of $f$.
2. $f'(a)$: how fast it is changing at $a$. Units of $f$ per unit of input.

They are independent. A quantity can be large and shrinking, small and growing, or anything else. If $N(t)$ is a fish population with $N(10) = 430$ and $N'(10) = -12$, there are 430 fish on day 10 and the population is falling at about 12 fish per day at that moment. There is no such thing as $-12$ fish.

The same independence holds one level up. $f'$ and $f''$ describe different things, so all four sign combinations occur:

1. $f' > 0$ and $f'' > 0$: growing, and growing faster.
2. $f' > 0$ and $f'' < 0$: still growing, but the gains are shrinking.
3. $f' < 0$ and $f'' > 0$: still falling, but the losses are shrinking.
4. $f' < 0$ and $f'' < 0$: falling, and falling faster.

A negative $f''$ never by itself means the quantity is going down. It means the rate is going down, which is a statement about $f'$.

The sign of $f'(a)$ carries exactly one piece of information: which way $f$ is moving at $a$. Three things it does not carry:

1. The sign of $f$ itself. An altitude of 340 meters with $A'(5) = -8$ is a balloon that is high up and coming down, not a balloon underground.
2. A total change. $W'(4) = -1.5$ pounds per week does not mean 1.5 pounds have been lost. It is the rate at that instant, and totals are Unit 6 work.
3. A guarantee about later. A rate at $t = 4$ describes $t = 4$. If it holds long enough it predicts, which is what Topic 4.6 makes precise.

One more habit worth building now: rates convert like any other unit. A pool gaining $6.4$ gallons per minute is gaining $6.4 \times 60 = 384$ gallons per hour at that instant. Multiplying, not dividing, because the time unit got bigger and more minutes fit inside it.