4.1 The Derivative: A Ratio at One Moment
The problem
A chemist follows the disappearance of a reactant A and fits the concentration to [\mathrm{A}](t) = 0.80\,e^{-0.139\,t} \qquad ([\mathrm{A}] \text{ in M},\; t \text{ in minutes}). Her kinetics textbook says that the rate at which A disappears is proportional to how much A is present right now. Notice the phrase. A rate law is a statement about a single instant: it ties the speed at this moment to the concentration at this moment. But every rate we have computed so far was a rate over an interval — “0.60 M lost in ten minutes,” or the slope of a calibration line between two standards. How does one extract a speed at one moment from a function that only gives concentrations at moments?
The question has the shape of a trap, and it is worth seeing the trap clearly before escaping it. A speed is a change divided by an elapsed time. At a single instant, no time elapses and nothing changes: the ratio is zero divided by zero. The rest of this section is the story of how that meaningless ratio becomes the most useful quantity in physical chemistry.
What an average rate is, and what is wrong with it
Pick any two times, t and t+h. The average rate of change of [\mathrm{A}] over that interval is \frac{[\mathrm{A}](t+h)-[\mathrm{A}](t)}{h} \qquad (\text{units: M/min}). Geometrically it is the slope of the secant line through the two points on the graph. This is a perfectly good physical quantity — but ask it for the rate at t = 0 and it gives an unsettling answer. Starting at t=0 and looking forward five minutes gives -0.080 M/min; looking forward one minute gives -0.104 M/min; looking forward a tenth of a minute gives -0.110 M/min. Three different answers for the same moment.
The reason is that each answer quietly contains a choice you made: the length h of the interval. An average rate is a property of the interval you picked, not of the chemistry at the moment you care about. What we want is the average rate with that arbitrary choice removed.
Shrinking the interval: the ratio settles
If the choice of h is the problem, let it shrink and watch what the quotient does. The table follows [\mathrm{A}](t) = 0.80\,e^{-0.139t} from t = 0.
| h (min) | change in [\mathrm{A}] (M) | quotient \Delta[\mathrm{A}]/h (M/min) |
|---|---|---|
| 5 | -0.40074 | -0.08015 |
| 1 | -0.10382 | -0.10382 |
| 0.1 | -0.01104 | -0.11043 |
| 0.01 | -0.00111 | -0.11112 |
| 0.001 | -0.00011 | -0.11119 |
Look at the middle column and then at the last one, because the whole idea lives in the contrast between them. The change in concentration is collapsing toward zero — by a factor of ten each row — and so is the elapsed time h. Both parts of the ratio are vanishing. Yet the ratio does not vanish; it creeps toward a definite number, -0.1112 M/min. A ratio of two vanishing quantities need not vanish. When numerator and denominator shrink together in step, the ratio can settle. That settled value — which no single interval produces but which every short enough interval approximates — is what “the rate at an instant” has to mean.
Remark. Press play in the derivative studio and watch both quantities collapse while the quotient locks onto its limit. Then move the anchor time t_0: the limit changes with the moment you ask about, so the instantaneous rate is itself a function of time.
Making it a definition
The table suggests a definition, and the symbols are worth reading in words.
Definition 4.1.1 (derivative). For a function f and a point x, the derivative of f at x is the number the difference quotient settles to as the interval shrinks: f'(x) = \lim_{h \to 0}\; \frac{f(x+h) - f(x)}{h}, whenever that limit exists. In Leibniz’s notation the same number is written \dfrac{df}{dx}.
Three readings of \dfrac{df}{dx} are useful and should be held together. It is the instantaneous rate: how fast f is changing per unit x at this moment. It is the slope of the tangent line: the secant line pivots toward the tangent as h \to 0. And it is the best straight-line approximation: near x, the function behaves like the line through (x, f(x)) with slope f'(x) — the very reason that the shortcut e^{x} \approx 1 + x you used on the exam works near zero. Notice too what the symbol is not: \frac{df}{dx} is not a fraction of two tiny numbers to be cancelled. It is the name of the settled value of a ratio, and its units are always (units of output) per (unit of input) — here M/min.
Why you cannot just set h = 0 — and what actually happens
If the quotient is 0/0 at h = 0, how can its value at h near zero be meaningful? Because the algebra can often remove the problem before h goes to zero. The common factor that makes the numerator vanish is also in the denominator; cancel it while h \neq 0, and what is left has no difficulty at h = 0.
Take the simplest nontrivial function, f(t) = t^{2}: \frac{(t+h)^{2} - t^{2}}{h} = \frac{2th + h^{2}}{h} = 2t + h \;\longrightarrow\; 2t \quad (h \to 0). The numerator is not merely small: it is h times something (2t + h). That structure — numerator proportional to h — is exactly what lets the ratio survive the collapse.
Example 4.1.1 — a filling reactor, from the definition. The product amount in a batch reactor is n(t) = 3.0\,t - 0.25\,t^{2} (mol, t in hours). Find n'(t) and then the rate at t = 2 h.
Setting it up. The derivative is the settled value of [\,n(t+h) - n(t)\,]/h. Everything hinges on expanding n(t+h) and spotting the common factor.
Solution. n(t+h) - n(t) = 3.0h - 0.25\,(2th + h^{2}) = h\,(3.0 - 0.50\,t - 0.25\,h). Dividing by h: \frac{n(t+h)-n(t)}{h} = 3.0 - 0.50\,t - 0.25\,h \;\longrightarrow\; n'(t) = 3.0 - 0.50\,t \quad (h \to 0). At t = 2 h: n'(2) = 2.0 mol/h.
Check. At t = 2 the quotient is 2.0 - 0.25\,h, so the numerical values for h = 2,\, 1,\, 0.5,\, 0.1 are 1.50,\, 1.75,\, 1.875,\, 1.975 mol/h, creeping to 2.0 exactly as the formula says. Units: mol per hour. Limiting case: at t = 0 the rate is 3.0 mol/h, the initial rate, and it falls linearly as the reactor depletes its reagent — the sign and trend a filling-then-saturating process should have.
The exponential: where first-order kinetics comes from
Now return to the chemist’s function. Writing [\mathrm{A}](t+h) = 0.80\,e^{-0.139t}\,e^{-0.139h}, the quotient becomes \frac{[\mathrm{A}](t+h)-[\mathrm{A}](t)}{h} = 0.80\,e^{-0.139\,t}\cdot\frac{e^{-0.139\,h} - 1}{h} = [\mathrm{A}](t)\cdot\frac{e^{-0.139\,h} - 1}{h}. Something remarkable has happened: the whole dependence on the moment t has factored out as [\mathrm{A}](t), and all the h-dependence sits in one clean factor. That factor settles as h shrinks:
| h | 1 | 0.1 | 0.01 | 0.001 |
|---|---|---|---|---|
| \dfrac{e^{-0.139h}-1}{h} | -0.12977 | -0.13804 | -0.13890 | -0.13899 |
— toward -0.139. Therefore \frac{d[\mathrm{A}]}{dt} = -0.139\,[\mathrm{A}](t). This is the chemical payoff of the section. The statement “the rate of disappearance is proportional to the amount present” is not an extra physical assumption stacked on top of the exponential: it is what the exponential is, read through the definition of the derivative. Conversely, a first-order rate law -d[\mathrm{A}]/dt = k[\mathrm{A}] forces the concentration to decay exponentially. At t = 0 the rate is -0.139 \times 0.80 = -0.1112 M/min — the number the table was approaching — and at t = 10 min, with 0.199 M left, it has fallen to -0.0277 M/min. The same curve has a different slope at every moment, which is why the derivative is itself a function.
Remarks and cautions
An average rate is not an instantaneous rate. Reporting -0.080 M/min as “the rate at t=0” because you measured over five minutes is the secant-for-tangent error. The two agree only when the curve is straight (a zero-order reaction), and the gap between them is exactly what shrinking h closes.
Units come with the derivative. d[\mathrm{A}]/dt is in M/min, not M. A derivative reported without its “per-input” unit is incomplete — and, as the next section will show, a missing unit is the quickest way to catch a dropped factor.
Remark. The limit must come out the same whether h shrinks through positive values (looking forward) or negative values (looking back). At a sharp corner — the kink in a heating curve at a phase boundary, say — the two disagree and no derivative exists there. Smooth chemical functions do not have this problem.
Summary
A rate over an interval is a ratio, and the interval length is an arbitrary choice that contaminates the answer; the rate at an instant is that same ratio with the choice removed. Shrinking the interval makes numerator and denominator both vanish, yet the ratio settles to a definite number whenever the numerator is proportional to h — the algebraic fact that lets us cancel h before letting it go to zero. That settled value is the derivative, readable three ways: as an instantaneous rate, as a tangent slope, and as the best straight-line approximation, with units of output per input. For the exponential decay [\mathrm{A}] = 0.80\,e^{-0.139t} the whole moment dependence factors out and the derivative is -0.139\,[\mathrm{A}] — first-order kinetics, derived rather than assumed. What the limit definition does not give us is speed: each function required its own expansion and its own cancellation. The next section asks whether chemistry’s functions are built from a few pieces whose derivatives we can establish once and reuse forever.
Practice until it sticks
Problems 1–3 and 7 of Problem Set 6 are the secant-to-tangent exercises for this section; the animated version is in the derivative studio.