The derivative — capturing "how fast, right now?"
The Problem. You are running a kinetics experiment. A reactant A is being consumed, and every ten seconds the spectrometer logs its concentration: 0.500 M at the start, 0.409 M at ten seconds, 0.335 M at twenty, 0.274 M at thirty. Your advisor looks over your shoulder and asks a question that sounds innocent: "What's the rate at t = 25 seconds?"
You pause. You have no measurement at t = 25 — the instrument recorded at 20 and at 30. But the difficulty runs deeper than a missing data point. Even if you had a reading at exactly 25 seconds, a single concentration is not a rate. A rate is about change, and change needs two points to compare. Yet the moment you pick two points, you are describing an interval — and your advisor asked about an instant. The reaction is slowing down the whole time, so the rate at t = 25 is genuinely different from the rate at t = 20 or t = 30. There seems to be no honest way to answer.
Somehow you must extract "how fast, right now" from data that only ever tells you "how much, at this time." This chapter builds the tool that does exactly that — the derivative — and by the end we will answer the advisor's question two independent ways and watch the answers agree.
Before attacking the advisor's problem, it is worth noticing how often this exact situation arises. The question "how fast?" is arguably the most common quantitative question in all of chemistry, and it always has the same hidden structure: some quantity depends on time (or on position, or on temperature), and we want to know how rapidly it is changing at one particular moment. Consider how many familiar measurements are secretly this question in disguise:
| Question | What you're really asking |
|---|---|
| "What's the reaction rate?" | How fast is concentration changing? |
| "What's the current?" | How fast is charge flowing? |
| "What's the velocity?" | How fast is position changing? |
| "What's the heat flow?" | How fast is energy transferring? |
Every row of this table names a different instrument and a different laboratory, but mathematically they are one question. In each case there is a function — concentration as a function of time, charge as a function of time, energy as a function of time — and what we are asking for is the instantaneous rate of change of that function. If we can solve the problem once, in the abstract, we solve it everywhere at once. That is the economy of mathematics: one tool, built carefully, serves every column of the table. The tool is the derivative, and to build it we should first understand precisely why the obvious approach falls short.
The obvious first answer to the advisor is to take the two measurements that bracket t = 25 and divide the change in concentration by the change in time. Between t = 20 s and t = 30 s the concentration fell from 0.335 M to 0.274 M, a drop of 0.061 M over 10 seconds. So the concentration is changing at about −0.0061 M per second, and we might report a rate of roughly 6.1 × 10⁻³ M/s. This quantity has a name and a precise definition:
$$\text{Average rate} = \frac{f(b) - f(a)}{b - a} = \frac{\Delta f}{\Delta x}$$
This is the slope of the secant line connecting (a, f(a)) and (b, f(b)).
For many purposes the average rate is good enough, and it has the great virtue of being computable directly from data. But notice what we had to give up to compute it. The average rate is a property of an interval, not of a moment. It tells us what happened between 20 and 30 seconds, smearing together the faster chemistry near the start of the interval with the slower chemistry near the end. Our advisor asked about t = 25 exactly. If the rate is changing within the interval — and in any real reaction it is — then no choice of interval gives us precisely what was asked.
The natural response is to shrink the interval. An average over [24, 26] would smear less than an average over [20, 30]; an average over [24.9, 25.1] would smear less still. Geometrically, each shrinking of the interval tilts the secant line — the straight line through the two endpoints — closer and closer to a limiting position: the tangent line, the straight line that just grazes the curve at t = 25 and matches its steepness there. The slope of that tangent line is the number we are after. The interactive below lets you watch this convergence happen.
Before you touch the slider: the secant slope and the true derivative are displayed below the graph, along with their difference. Predict — as you drag h toward zero, does the error shrink steadily, or does it suddenly jump to zero at some point? Decide, then test.
You should have seen the error shrink gradually — there is no magic value of h where the secant suddenly becomes the tangent. At h = 0.05, the smallest the slider allows, the secant slope is close to the true derivative but still not equal to it. This observation is the crux of the whole chapter: no single secant line ever gives the exact answer, yet the sequence of secant slopes is clearly heading somewhere. The question is how to lay hands on the destination.
Why not simply finish the job and set h = 0? Because the average rate is a ratio, $\frac{f(a+h) - f(a)}{h}$, and at h = 0 both the numerator and the denominator vanish. The expression becomes 0/0, which is not a number — it is the absence of an answer. An interval of width zero contains no change to measure. This is the impasse: we need the interval to be zero, and we cannot allow the interval to be zero.
The escape is the idea from Lecture 11: the limit. Instead of asking "what is the ratio at h = 0?" — a question with no answer — we ask "what value does the ratio approach as h shrinks toward 0?" That is a different question, and it does have an answer. You watched it have an answer in the interactive above: as h marched toward zero, the secant slope marched toward a definite number and stabilized. The limit names that number without ever requiring us to divide by zero. The derivative is defined as exactly this limit:
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
if this limit exists.
Read the definition from the inside out, because every piece of it earns its place. The inner fraction $\frac{f(a+h)-f(a)}{h}$ is just the average rate over the small interval from a to a + h — the secant slope, the thing we can always compute. The limit then asks where those secant slopes are heading as the interval shrinks. The result, f′(a), is a single number attached to the single point a: the instantaneous rate of change, the slope of the tangent line, the answer to the advisor's kind of question. Note also the caveat "if this limit exists" — it is not decoration. For some functions, at some points, the secant slopes refuse to settle on any destination, and then there simply is no derivative there. Section 8 shows what that failure looks like.
The derivative was invented more or less simultaneously by Newton and Leibniz, and their notational descendants both survive. You will need to read all of the following fluently, because chemistry texts switch between them without warning:
| Notation | Read as | Origin |
|---|---|---|
| f'(x) | "f prime of x" | Lagrange |
| df/dx | "dee f dee x" | Leibniz |
| d/dx f(x) | "dee dee x of f" | Operator form |
| ḟ | "f dot" | Newton (time) |
Lagrange's prime notation f′(x) is the most compact, and we use it when the variable is obvious. But Leibniz's notation df/dx is the one chemists live with, and it deserves a moment of reflection. It is built to look like a fraction — an infinitesimally small change in f divided by an infinitesimally small change in x — because that is how Leibniz thought of it. Strictly speaking it is not a fraction; it is the limit of fractions, a single indivisible symbol. But the notation is engineered so well that treating it like a fraction usually gives correct results, and in Lecture 19 we will exploit this shamelessly when separating variables in differential equations. A second virtue: df/dx displays its own units. If f is a concentration in M and x is a time in seconds, then df/dx is in M/s — the units of the numerator divided by the units of the denominator. A derivative always carries units, and checking them is the fastest way to catch an error.
The definition is not only a conceptual statement — it is a recipe you can actually run. The pattern is always the same three moves: write the difference quotient, do algebra until the troublesome h in the denominator cancels away, and then let h → 0 in what remains. Watch the pattern work on the simplest interesting function, f(x) = x².
First, the difference quotient. We need f(x + h), which is (x + h)² = x² + 2xh + h², so:
$$\frac{f(x+h) - f(x)}{h} = \frac{(x^2 + 2xh + h^2) - x^2}{h} = \frac{2xh + h^2}{h}$$
Here is the crucial move. The x² terms cancel — the numerator keeps only terms that contain h. That is no accident: f(x+h) − f(x) measures the change in f, and the change must vanish as h does. Because every surviving term carries a factor of h, we may divide it out:
$$\frac{2xh + h^2}{h} = 2x + h$$
And now the expression is harmless at h = 0. The division by zero that blocked us in Section 3 has been algebraically dissolved, and the limit is found by simply letting h go:
$$f'(x) = \lim_{h \to 0}\,(2x + h) = 2x$$
Pause on what kind of object we obtained. We asked for the slope at a point and received a function: f′(x) = 2x hands us the slope at every point at once. At x = 1 the parabola climbs with slope 2; at x = 3 it climbs steeply with slope 6; at x = −2 it descends with slope −4; and at x = 0, the bottom of the bowl, the slope is exactly zero. One computation, infinitely many tangent lines.
Problem. Find f′(x) for f(x) = 1/x.
Setting it up. Same recipe: difference quotient, algebra until h cancels, then h → 0. The new wrinkle is that f(x+h) − f(x) is a difference of fractions, so the algebra step will be "combine over a common denominator" rather than "expand a square." We should expect trouble only where the function itself is in trouble, namely x = 0.
Solution. The difference quotient is:
$$\frac{1}{h}\left(\frac{1}{x+h} - \frac{1}{x}\right) = \frac{1}{h}\cdot\frac{x - (x+h)}{x(x+h)} = \frac{1}{h}\cdot\frac{-h}{x(x+h)} = \frac{-1}{x(x+h)}$$
Once again the change in f produced a factor of h in the numerator, and dividing by h removed it. Letting h → 0:
$$f'(x) = \lim_{h\to 0} \frac{-1}{x(x+h)} = -\frac{1}{x^2}$$
Check. The sign is right: for x > 0, the function 1/x is falling, and our derivative is negative everywhere. The behavior is right: near x = 0 the hyperbola plunges almost vertically and indeed −1/x² blows up there; far from the origin the curve flattens and −1/x² duly shrinks toward zero. And the answer agrees with the power rule of Lecture 13 — write 1/x as x⁻¹, and nxⁿ⁻¹ gives −x⁻² — so the general rule we adopt later is consistent with the definition we sweated through here.
One more computation from the definition, because its result organizes half of physical chemistry. Take f(x) = eˣ. The difference quotient factors cleanly, since e^(x+h) = eˣ·eʰ:
$$f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h}$$
Everything depending on x has been pulled outside the limit; what remains inside is a pure number, the slope of eˣ at the origin. For the base e — and this is precisely what makes e the natural base, the reason it is worth singling out 2.71828… from all possible bases — that limiting value is exactly 1. Therefore:
$$f'(x) = e^x$$
The exponential function is its own derivative: at every point, its slope equals its height. This is why eˣ appears wherever a quantity changes at a rate proportional to itself — first-order kinetics, radioactive decay, the Boltzmann distribution, RC circuits. Nature keeps posing the equation "rate of change ∝ current amount," and eˣ is its built-in solution. We will lean on this constantly, starting in Section 6.
Before the next interactive: it plots f(x) = x² on the left and f′(x) = 2x on the right, with a movable point. Predict — when the point sits exactly at the bottom of the parabola, where is the corresponding point on the right-hand graph? And as you slide from left of the minimum to right of it, what single feature of the right-hand graph records the moment you pass the bottom? Decide, then drag.
You should have found that the bottom of the parabola corresponds to the point where the right-hand graph crosses zero. That is the debrief in one sentence: the derivative converts a geometric feature (the bottom of a valley) into an algebraic one (f′ = 0), which a computer — or a tired student during an exam — can find without ever drawing the curve. Notice too that f′ is negative everywhere the parabola descends and positive everywhere it climbs; the sign of the derivative is a running commentary on the function's behavior.
It is worth collecting these readings of f′ in one place, because using the derivative fluently means translating between them without conscious effort. Geometrically, the sign of f′(a) classifies the local behavior: where f′(a) > 0 the function is increasing — the tangent line tilts uphill; where f′(a) < 0 it is decreasing; and where f′(a) = 0 the tangent is horizontal, marking a point where the function has momentarily stopped changing. These zero-slope points are where maxima and minima hide, which is why Lecture 14 will turn "find the extremum" into "solve f′ = 0." A word of caution that we will sharpen there: f′ = 0 marks a candidate extremum, not a guaranteed one — the function f(x) = x³ has zero slope at the origin yet keeps right on climbing through it.
Physically, the derivative of any quantity with a name usually has a name of its own. Many quantities you have treated as independent concepts are secretly derivative pairs:
| If f(t) represents... | Then f'(t) is... |
|---|---|
| Position | Velocity |
| Velocity | Acceleration |
| Concentration | Reaction rate |
| Charge | Current |
| Energy | Power |
Each row is the same mathematical relationship wearing different laboratory clothes. Velocity does not merely relate to position — it is the derivative of position with respect to time; current is the derivative of charge. Once you see the pattern, the units become self-explanatory: every derivative inherits the units "numerator per denominator," so concentration in M differentiated with respect to time in s must yield M/s, and energy in J differentiated with respect to time yields J/s — which we long ago renamed the watt.
There is a third reading of the derivative, alongside the geometric and the physical, that working scientists use most of all: the derivative is the function's best linear stand-in. The tangent line at a point hugs the curve so closely that, for small excursions h away from the point, the line is an excellent substitute for the function itself:
$$f(a + h) \approx f(a) + f'(a)\,h$$
Read it as a recipe: new value ≈ old value plus (rate of change) × (how far you moved). This linear approximation is how you estimate without computing — e^0.1 must be close to 1 + 0.1 = 1.1, because eˣ has value 1 and slope 1 at the origin (the true value is 1.105). It is also the first word of a longer story: keeping more terms beyond the linear one yields the Taylor series of Lecture 16, and the harmonic approximation that underlies all of vibrational spectroscopy is exactly this idea applied to a potential well. For now, internalize the slogan: zoom in far enough on any differentiable function, and it looks like a straight line whose slope is the derivative.
Now we can give the advisor's question a precise meaning. For a reaction A → Products, the concentration [A] is a function of time, and the instantaneous rate of the reaction is defined through its derivative:
$$\text{Rate} = -\frac{d[A]}{dt}$$
The minus sign deserves a sentence rather than a footnote, because dropped minus signs are the leading cause of death on kinetics exams. Since A is being consumed, [A] is a decreasing function, so its derivative d[A]/dt is negative. But chemists insist, by convention, that a reaction proceeding forward has a positive rate. The minus sign reconciles the two: it flips the negative derivative into the positive number we report. If we instead tracked a product P, its concentration grows, d[P]/dt is already positive, and the rate is +d[P]/dt with no sign correction. (When stoichiometry is not one-to-one — say 2A → P — one also divides by the stoichiometric coefficient, defining Rate = −(1/2) d[A]/dt, so that every species in the reaction reports the same rate. For this chapter, our reactions are all A → Products and the coefficient is 1.)
Problem. Using the data from the opening — [A] = 0.500, 0.409, 0.335, 0.274 M at t = 0, 10, 20, 30 s — find the rate at t = 25 s.
Setting it up. We will answer twice, by two independent routes, and let the agreement between them be our verification. Route one uses only the raw data: the best estimate of an instantaneous rate from discrete measurements is the average rate over the smallest interval centered on the time of interest — here, [20 s, 30 s], whose midpoint is exactly 25 s. Route two uses a model: this data is consistent with first-order decay, [A](t) = (0.500 M)·e^(−kt) with k = 0.020 s⁻¹ (check: at t = 10, 0.500·e^(−0.2) = 0.409 ✓), and a model can be differentiated exactly.
Solution, route one (from data). The centered average rate is:
$$\frac{\Delta[A]}{\Delta t} = \frac{0.274 - 0.335}{30 - 20} = \frac{-0.061}{10} = -6.1 \times 10^{-3}\ \text{M/s}$$
so Rate ≈ 6.1 × 10⁻³ M/s.
Solution, route two (from the model). Differentiating the exponential — using d(eˣ)/dx = eˣ from Section 4 together with the chain rule we will justify properly in Lecture 13 — gives d[A]/dt = −k·(0.500)·e^(−kt). At t = 25 s:
$$-\frac{d[A]}{dt}\bigg|_{t=25} = (0.020)(0.500)\,e^{-0.5} = (0.010)(0.6065) = 6.07 \times 10^{-3}\ \text{M/s}$$
Check. Units first: M/s on both routes, as a rate of concentration change must be. Sign: both rates come out positive after the convention's minus sign, as a forward reaction demands. And the two routes agree to within half a percent — 6.1 versus 6.07 × 10⁻³ M/s — which tells us the centered difference over a 10-second window was an excellent stand-in for the true derivative here. The agreement is not luck: the centered secant errs symmetrically, overestimating on one side of t = 25 what it underestimates on the other. This, finally, is the honest answer to the advisor — and the reasoning behind it is the entire content of this chapter.
Before the next interactive: it plots first-order decay with adjustable k and a movable time marker, displaying both the measured slope −d[A]/dt and the quantity k[A]. Predict — if you double k, does the rate at t = 10 s double as well, or change by some other factor? (Careful: k appears twice, once in front and once in the exponent.) Decide, then test.
The debrief: doubling k less than doubles the rate at t = 10 s, because the larger k has already consumed more reactant by then — the prefactor doubles, but e^(−kt) shrinks. More importantly, compare the two displayed numbers as you slide: −d[A]/dt and k[A] track each other exactly, at every k and every t. That identity, Rate = k[A], is the defining equation of first-order kinetics, and you should now read it with new eyes. It says the derivative of the concentration is proportional to the concentration itself — which, as Section 4 promised, is precisely the situation that exponentials are born to describe. The decay curve is exponential because the rate is proportional to the amount; in Lecture 19 we will run this logic forward and derive the curve from the rate law.
The derivative's domain is not only time. Whenever a quantity varies with position, its derivative with respect to position is equally meaningful, and the most important example in chemistry is the potential energy V(x) — for instance, the energy of a diatomic molecule as a function of its bond length. The relationship between energy and force is:
$$F = -\frac{dV}{dx}$$
To feel why this is true rather than memorize it, picture a ball resting on a hillside whose height profile is V(x). The ball rolls downhill — always toward lower potential energy. Where the hill slopes upward to the right (dV/dx > 0), "downhill" is to the left, so the force must be negative; where the hill slopes downward to the right (dV/dx < 0), the ball is pushed rightward and the force is positive. In both cases the force is opposite in sign to the slope, which is exactly what the minus sign encodes: force pushes down the energy gradient. The magnitude matters too — a steep slope means a large derivative and hence a strong force, a gentle slope a weak one. And where the slope vanishes, dV/dx = 0, there is no force at all: the system is at equilibrium. A bond sits at the bottom of its potential well precisely because that is where the derivative of the energy is zero.
The harmonic potential V = ½kx² makes this concrete. Differentiating (by the power rule pattern you saw for x²) gives dV/dx = kx, so F = −kx: a restoring force proportional to displacement, pointing back toward the origin. You have met this before as Hooke's law — and now you can see it is not an independent law at all, but the derivative of a parabolic energy. This is the mathematical heart of why molecules vibrate: near the bottom of any smooth well, the potential looks parabolic, so the force looks like Hooke's law, so the motion is oscillation.
Before the next interactive: select the double-well potential, V = x⁴ − 2x². Predict — at how many positions is the force exactly zero, and at which of them would a particle, nudged slightly, return rather than run away? Decide, then explore with the slider.
You should have found three equilibria: the two well bottoms at x = ±1, where a nudged particle rolls back (stable), and the central hump at x = 0, where the force is also zero but any nudge sends the particle accelerating away (unstable). Both kinds satisfy dV/dx = 0; what distinguishes them is whether the slope around the equilibrium pushes back or pushes away — a question about how the derivative itself is changing, which is the business of the second derivative and of Lecture 14. Double wells like this one are not a textbook curiosity: they are the energy landscape of any molecule with two stable conformations, and the hump between the wells is the barrier that reaction rates fight against.
It would be a mistake to file the derivative away as a chemistry tool, or even a science tool. The phrase "rate of change" belongs to no single discipline, and the moment a field starts measuring how one quantity responds to another, it has — knowingly or not — started doing calculus. The economist who asks how much an extra unit of production costs, the ecologist who asks how fast a population is expanding, the pharmacologist who asks how quickly a drug clears the bloodstream: each is computing a derivative, and each gives it a different name. This is the central claim of the whole course made concrete — mathematics and its applications are the same thought wearing different clothes.
Consider three questions from three faculties, set beside the chemistry we just did:
| Field | The quantity | Its derivative | What the field calls it |
|---|---|---|---|
| Chemistry | Concentration [A](t) | d[A]/dt | Reaction rate |
| Economics | Total cost C(q) | dC/dq | Marginal cost |
| Ecology | Population P(t) | dP/dt | Growth rate |
| Pharmacology | Plasma drug level C(t) | dC/dt | Elimination rate |
Notice that one row differs in spirit. In economics the independent variable is not time but quantity produced — marginal cost is dC/dq, the slope of total cost against output. The derivative does not care; "per unit of the input" is its only requirement, whether that input is measured in seconds or in units manufactured. The interactive below plots all four cases in a single frame. Switch fields and watch the curve, the units, and the name change while the mathematical object — the slope of the tangent line — stays exactly the same.
Before you switch fields: the chemistry and pharmacology curves both fall, while the economics and ecology curves both rise. Predict — for which of the four will the displayed derivative be negative? And for the S-shaped ecology curve, predict where along it the growth rate is largest: at the start, in the middle, or near the top. Decide, then explore.
The debrief: the two falling curves — chemistry and pharmacology — show negative derivatives, because the quantity is being consumed or cleared; the convention of attaching a minus sign (Section 6) is what lets both fields report a positive "rate." The two rising curves show positive derivatives. And the ecology curve makes a point a straight line never could: its growth rate is not largest at the start (few individuals reproducing) nor near the top (crowding, scarce resources), but exactly in the middle, at the steepest part of the S — the inflection point, where the population is half its carrying capacity. We meet this curve again, and prove the half-capacity claim, when logistic growth returns as a differential equation in Lecture 19.
Problem. A small fermentation plant's total cost of producing q liters of ethanol per day is C(q) = 200 + 8q + 0.05q² dollars. What is the marginal cost at q = 100 L/day, and what does it mean?
Setting it up. "Marginal cost" is the economist's name for dC/dq — the rate at which total cost rises as output rises. The function is a polynomial, so we differentiate term by term using the power-rule pattern from Section 4 (a constant differentiates to 0, the linear term to its coefficient, and 0.05q² to 0.10q). Marginal cost answers a decision: is the next liter worth making?
Solution. Differentiate:
$$\frac{dC}{dq} = 0 + 8 + 0.10\,q$$
Evaluate at q = 100: dC/dq = 8 + 0.10(100) = $18 per liter.
Check. Units: dollars per liter — a cost per unit of output, exactly what "marginal" should be. Magnitude by brute force: C(100) = 200 + 800 + 500 = $1500, and C(101) = 200 + 808 + 0.05(10201) = $1518.05, a difference of $18.05. The derivative's $18 matches this "one more unit" cost to within a nickel — the small discrepancy is the curvature the tangent line ignores, the same approximation error you watched shrink in Section 2. The marginal cost rises with q (the 0.10q term), which is the economics of strained capacity: the more you already produce, the more the next liter costs.
Problem. A population under limited resources follows logistic growth, whose rate is governed by dP/dt = rP(1 − P/K), with intrinsic rate r and carrying capacity K. At what population size is the growth rate dP/dt largest?
Setting it up. Here the derivative dP/dt is itself given to us as a function of P, and we are asked to maximize it. A quantity is largest where its rate of change vanishes — so we differentiate the growth rate with respect to P and set it to zero. This is a derivative of a derivative, a preview of Lecture 14, but the algebra is just the product/power rule on a quadratic in P.
Solution. Expand the growth rate: dP/dt = rP − (r/K)P². Differentiate this with respect to P:
$$\frac{d}{dP}\!\left(rP - \frac{r}{K}P^2\right) = r - \frac{2r}{K}P$$
Set equal to zero: r − (2r/K)P = 0, which gives P = K/2, half the carrying capacity.
Check. The result is independent of r, as it should be — r sets how fast, not where the peak sits. At the endpoints the answer is sensible: at P = 0 there is no one to reproduce (rate 0), and at P = K the environment is full (rate 0), so the maximum must lie strictly between — and symmetry of the downward parabola rP(1−P/K) puts it dead center, at K/2. This matches what you saw on the S-curve in the interactive: the steepest point was the middle. Conservation biologists use exactly this fact — a fishery or harvested herd yields the most sustainable surplus when held near half its carrying capacity.
These are not contrived analogies. The relationships above are the working content of textbooks in our own library — Microeconomics and Macroeconomics, the Environmental Science of Freedman, the Principles of Pharmacology — where marginal cost, logistic growth, and drug half-life are introduced, often without ever using the word "derivative." You now hold the single tool beneath all of them.
The definition in Section 3 ended with a quiet condition — "if this limit exists" — and we owe it an honest look. The limit defining f′(a) can fail, and when it fails the function simply has no derivative at that point, no matter how much we might want one. The failure has a clear geometric signature. Approaching the point from the left, the secant slopes settle toward one value; approaching from the right, toward another; the derivative exists only when the two sides agree on a single, finite answer. The classic offender is f(x) = |x| at the origin: from the left every secant has slope −1, from the right every secant has slope +1, and no amount of shrinking h reconciles them. The graph has a corner, and a corner has no single tangent line.
Before the interactive: it offers three different failures — a corner, a cusp, and a vertical tangent. Predict which of the three has left and right secant slopes that do agree with each other, yet still fails to have a derivative. Then check yourself.
The answer is the vertical tangent, x^(1/3): both one-sided slopes head to +∞ together, agreeing perfectly — but +∞ is not a number, and the definition demands a finite limit. Agreement is necessary, not sufficient. One general fact is worth carrying away from these pathologies: differentiability implies continuity, but not the reverse. If a function has a derivative at a point, it cannot jump there — you cannot draw a tangent line to a gap. But the converse fails, and |x| is the standing counterexample: perfectly continuous through the origin, yet not differentiable there. Smoothness is a strictly stronger demand than unbrokenness. Nor are these failures merely mathematical curiosities for a chemist: potential energy surfaces develop cusp-like features where electronic states cross (the conical intersections of photochemistry), and at such points the force, −dV/dx, is genuinely ill-defined — the molecule's fate becomes a question the classical picture cannot answer.
Running the limit definition is illuminating, but nobody does it twice for the same function. Sections 4's computations — and others like them — get done once, in general, and recorded. The short table below covers the functions that account for nearly every derivative in this course; Lecture 13 adds the combination rules (product, quotient, chain) that let you differentiate anything built from these parts without ever touching a limit again.
| f(x) | f'(x) | Notes |
|---|---|---|
| c (constant) | 0 | Constants don't change |
| xⁿ | nxⁿ⁻¹ | Power rule |
| eˣ | eˣ | Its own derivative! |
| ln x | 1/x | |
| sin x | cos x | |
| cos x | -sin x |
Remark 1 — the sign of the rate. The most common kinetics error is writing Rate = d[A]/dt for a reactant. For anything being consumed, d[A]/dt is negative; the defined rate carries the compensating minus sign. A quick self-test before submitting any answer: a forward-running reaction must have a positive rate, and a derivative must have the sign of the actual trend (falling concentration ⇒ negative derivative). If your two signs don't reconcile through the definition, one of them is wrong.
Remark 2 — derivatives have units. f′ inherits the units of f divided by the units of the independent variable. A rate in M/s, a force in J/m (which is a newton), a heat capacity dU/dT in J/K. If a computed derivative comes out with impossible units, the setup was wrong before the calculus started. Checking units takes ten seconds and catches a remarkable fraction of errors.
Remark 3 — df/dx is not a fraction, but it plays one convincingly. The Leibniz symbol is a limit, a single object; the "df" and "dx" have no independent meaning at this stage. Treating them as separable quantities happens to work in many manipulations (and we will use it when solving differential equations), but it is a notational courtesy, not an algebraic right. When in doubt, fall back on the limit definition — it is the contract everything else must honor.
Remark 4 — every measured rate is an average. No instrument measures an instantaneous rate; every device averages over its response time, just as our spectrometer averaged over 10-second windows. The derivative is the idealization that these averages approach as the window shrinks. In practice this is a feature: as Worked Example 2 showed, a centered average over a modest window can approximate the true derivative astonishingly well. Knowing when it does — and when noisy data makes shrinking the window counterproductive — is the art of numerical differentiation, which we will revisit with error analysis in Lecture 22.
We began with a question the laboratory forces on every chemist — what is the rate right now? — and found that the obvious tool, the average rate Δf/Δx, answers a subtly different question about an interval rather than an instant. Shrinking the interval improved the answer but could never finish the job, because at zero width the ratio collapses to the meaningless 0/0. The limit broke the impasse: the derivative f′(a) is defined as the value the average rates approach, and it exists wherever the one-sided approaches agree on a finite number. Computed from the definition, the derivative of x² is 2x and the derivative of eˣ is itself — the second fact being the seed of all exponential behavior in chemistry. The derivative reads out geometrically as the tangent slope, physically as the instantaneous rate, and chemically as the reaction rate −d[A]/dt and the force −dV/dx; its zeros locate equilibria, and its sign tells you which way things are heading. The same object is the economist's marginal cost, the ecologist's growth rate, and the pharmacologist's elimination rate — one idea answering "how fast?" in every quantitative field. What it does not yet give us is an efficient calculus — running a limit for every new function would be unbearable — and supplying that machinery, the rules of differentiation, is the work of Lecture 13.
Use the limit definition to find f'(x) for:
A reaction has concentration profile [A](t) = 0.5e^(-0.1t) M.
d[A]/dt = -0.05e^(-0.1t), so Rate = -d[A]/dt = 0.05e^(-0.1t)
Check: units are M/s throughout; the rate is positive and decreasing with time, as a first-order decay demands; and Rate = k[A] holds at every answer (e.g., at t = 0: 0.1 × 0.5 = 0.05 ✓).
A particle moves in the potential V(x) = x⁴ - 2x².
Check: this is the double-well from Section 7's interactive — two stable wells at x = ±1 separated by an unstable hump at x = 0, exactly what you observed with the slider.
A biotech firm's total cost of producing q grams of a reagent is C(q) = 500 + 12q + 0.04q² dollars.
Check: units are dollars/g throughout; marginal cost rises with q, the signature of strained capacity; and the direct estimate C(51) − C(50) = $16.04 confirms part 2 to within the curvature term.
Early, resource-unlimited growth of a bacterial culture (or a human population) follows P(t) = P₀e^(rt). Take P₀ = 8.0 billion and r = 0.030 yr⁻¹.
Check: the doubling time depends only on r, not on P₀ — a hallmark of exponential growth, and the population analogue of a radioactive half-life. The same eˣ that decays in chemistry grows here; only the sign of the exponent differs.
After an injection, a drug's plasma concentration is C(t) = 80e^(−0.20t) mg/L, with t in hours.
−dC/dt = (0.20)(80)e^(−0.20t) = 16e^(−0.20t) mg/L/h
Check: units mg/L/h throughout; the elimination rate falls as the drug clears, just as a reaction rate falls as reactant is consumed. Pharmacologists call k the elimination rate constant — it is the chemist's rate constant under another name.
A simplified model of atmospheric CO₂ (in ppm) since 1960 is C(t) = 315 + 0.8t + 0.015t², where t is years after 1960. (This quadratic is a deliberate simplification of real Mauna Loa data, not an exact fit.)
Check: the model's outputs are physically sane — about 0.8 ppm/yr in the early 1960s and roughly 2.6 ppm/yr today both sit close to observed values, and C(60) = 315 + 48 + 54 = 417 ppm lands near the real 2020 figure. The positive second derivative is the mathematical content of the phrase "the problem is getting worse faster," and it is the same second-derivative idea that distinguished stable from unstable equilibria in Section 7.
The exercises above have fixed answers you can check against. This one is different: it watches how you answer. Get it right cleanly and the problems get harder; slip, and it works out which misconception you hit and nudges you toward the fix, handing you a fresh variant until it clicks. Clean-solve all three tiers — Easy, Medium, Hard — to master the skill.
References: McQuarrie, Mathematics for Physical Chemistry; Atkins, Physical Chemistry