The derivative has several names, and each is useful. The notation of Leibniz, $\frac{dy}{dx}$, also carries units, which makes it the language of rates in science. It is the same object as $f'(x)$ from The Derivative as a Function.
Key idea
If $y = f(x)$, these all mean the derivative: $f'(x)$, $y'$, $\dfrac{dy}{dx}$, $\dfrac{df}{dx}$, $\dfrac{d}{dx}f(x)$. The value at a point is $f'(a) = \left.\dfrac{dy}{dx}\right|_{x = a}$. The units of $\dfrac{dy}{dx}$ are units of $y$ per unit of $x$.
Why it works
$\frac{dy}{dx}$ comes from the difference quotient $\frac{\Delta y}{\Delta x}$: a change in $y$ divided by a change in $x$. A change in liters divided by a change in minutes is measured in liters per minute. Taking the limit as $\Delta x \to 0$ does not change the units. So the derivative of a quantity with respect to time is always "something per unit time": a rate.
| $y$ (units) | $x$ (units) | $\frac{dy}{dx}$ (units) |
|---|---|---|
| position (meters) | time (seconds) | meters per second (velocity) |
| cost (dollars) | items made | dollars per item (marginal cost) |
| temperature (°C) | depth (meters) | °C per meter |
| volume (liters) | time (minutes) | liters per minute (flow rate) |
Worked example: Interpret a derivative
$C(q)$ is the cost, in dollars, of making $q$ bicycles. What does $C'(100) = 85$ mean?
- $C'(100)$ is a rate at $q = 100$, in dollars per bicycle.
- At a production level of 100 bicycles, cost is rising at about \$85 per additional bicycle.
- So the 101st bicycle costs roughly \$85 more to make.
Answer: Making one more bicycle (after 100) adds about \$85 to the cost.
Worked example: Estimate with a rate
A tank holds $V(10) = 340$ liters at $t = 10$ minutes, and $V'(10) = -4$. Estimate $V(10.5)$.
- $V$ is falling at about 4 liters per minute near $t = 10$.
- In half a minute it changes by about $-4 \times 0.5 = -2$ liters.
- $V(10.5) \approx 340 - 2 = 338$ liters.
Answer: About 338 liters.
Common mistake
Giving the units of the derivative as the units of $y$, for example "$V'(10)$ is measured in liters." A derivative is a rate, so its units are always a ratio: liters per minute. Check: read $\frac{dV}{dt}$ aloud as "change in $V$ over change in $t$," and divide the units the same way.
Watch the difference between $\left.\frac{dy}{dx}\right|_{x=3}$ (differentiate, then substitute) and $\frac{d}{dx}\big[f(3)\big]$, which is the derivative of a constant, so it is 0.