4.2 Rules of Differentiation
The problem
The limit definition gave us a method, and Example 4.1.1 showed its price: expand, find the common factor, cancel, take the limit — for every function, from scratch. Chemistry does not hand us one function. It hands us the Arrhenius equation, Beer’s law, the ideal-gas law, the Michaelis–Menten curve, Coulomb’s law. If each needed its own expansion, differentiation would be a craft practiced by specialists.
It is not, because of a structural fact worth saying aloud: chemistry’s functions are built from a very small set of pieces, glued together in a very small set of ways. The pieces are powers x^{n}, exponentials e^{ax}, and logarithms \ln x. The ways of gluing are adding, scaling by a constant, multiplying, and nesting one function inside another. If we derive the derivative of each piece once, and the effect of each kind of gluing once, then every function in this chapter can be differentiated by reading it. Each rule below is therefore derived, not announced — because a rule you can rebuild is a rule you cannot misremember.
Gluing 1: sums and constant multiples — rates add
Two reactions run in the same flask and both release gas: n_1(t) and n_2(t) moles. The total is n = n_1 + n_2, and its rate of change should be the sum of the two rates. The definition says exactly that, because the difference quotient of a sum is the sum of the difference quotients: \frac{(n_1+n_2)(t+h)-(n_1+n_2)(t)}{h} = \frac{n_1(t+h)-n_1(t)}{h} + \frac{n_2(t+h)-n_2(t)}{h}. A constant factor passes straight through for the same reason — a detector with gain g reports g\,c(t), so its signal changes g times as fast as the concentration it tracks.
Theorem 4.2.1 (linearity). \;\dfrac{d}{dx}\big[a\,f(x) + b\,g(x)\big] = a\,f'(x) + b\,g'(x) for constants a, b.
Piece 1: powers
For f(x) = x^{n} the definition requires expanding (x+h)^{n}. For n = 2 we did it: x^2 + 2xh + h^2. For n = 3: (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3. The pattern is what matters. Subtract x^{n} and divide by h: the first term of the expansion cancels, the second survives as n\,x^{n-1} with its h divided out, and every remaining term still carries at least one factor of h — so it vanishes in the limit.
Theorem 4.2.2 (power rule). \;\dfrac{d}{dx}\,x^{n} = n\,x^{n-1}, for every real exponent n.
The exponent need not be a positive integer. Check n = -1 directly: \dfrac{1/(x+h) - 1/x}{h} = \dfrac{-h}{x(x+h)\,h} = \dfrac{-1}{x(x+h)} \to -\dfrac{1}{x^{2}}, which is the power rule with n = -1. Two laws of chemistry fall out immediately. The ideal-gas pressure at fixed n and T is P = nRT\,V^{-1}, so \dfrac{dP}{dV} = -\dfrac{nRT}{V^{2}}: the pressure falls as the volume grows, and falls faster when the gas is compressed. And the Coulomb energy of two charges, V(r) = kq_1q_2\,r^{-1}, gives a force F = -\dfrac{dV}{dr} = \dfrac{kq_1q_2}{r^{2}} — the inverse-square law is nothing but the derivative of an inverse-first-power energy.
Piece 2: the exponential
The exponential is the one function whose definition-quotient we already took apart: all dependence on the position factors out, leaving (e^{h}-1)/h, which settles to 1 — that is what the number e is. So \dfrac{d}{dx}\,e^{x} = e^{x}: the function whose rate of change at every point equals its own value. Section 4.1 showed the same structure with a rate constant in the exponent. The rule for a general exponent a needs the next kind of gluing.
Gluing 2: composition — rates multiply
[\mathrm{A}](t) = 0.80\,e^{-0.139t} is not a pure exponential in t; it is an exponential of a different quantity, u = -0.139\,t, which is itself a function of time. This is nesting, and it needs a rule. Think of it as a chain of cause and effect: time changes u at one rate, and u changes [\mathrm{A}] at another. If u changes -0.139 units per minute and [\mathrm{A}] changes by [\mathrm{A}] molar per unit of u, then [\mathrm{A}] changes by the product per minute.
Theorem 4.2.3 (chain rule). If y = F(u) and u = g(x), then \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}.
Why rates multiply. A small change \Delta x produces a change \Delta u \approx g'(x)\,\Delta x, which in turn produces \Delta y \approx F'(u)\,\Delta u = F'(u)\,g'(x)\,\Delta x. Divide by \Delta x and shrink: the two conversion factors multiply.
For the decaying reactant: \dfrac{dy}{du} = 0.80\,e^{u} = [\mathrm{A}] and \dfrac{du}{dt} = -0.139 min^{-1}, so \frac{d[\mathrm{A}]}{dt} = [\mathrm{A}] \cdot (-0.139\ \text{min}^{-1}) = -0.139\,[\mathrm{A}]. The units check this rule for free. \dfrac{dy}{du} is in M per (dimensionless u), \dfrac{du}{dt} is in min^{-1}, and their product is M/min — the only unit a rate of change of concentration can have. The factor -0.139 that the chain rule contributes is not optional decoration; it is the only place the “per minute” can come from. Generalizing, \dfrac{d}{dx}\,e^{ax} = a\,e^{ax}.
Example 4.2.1 — how sensitive is a rate constant to temperature? The Arrhenius equation is k(T) = A\,e^{-E_a/(RT)}. Find dk/dT, and for E_a = 75 kJ/mol at T = 298 K, how fast does k change per kelvin, as a fraction of itself?
Setting it up. The exponent is a function of T: u(T) = -\dfrac{E_a}{R}\,T^{-1}. Outer function A\,e^{u}; inner function u(T). Chain rule, with the power rule inside.
Solution. \dfrac{du}{dT} = +\dfrac{E_a}{R}\,T^{-2} = \dfrac{E_a}{RT^{2}} and \dfrac{dk}{du} = A\,e^{u} = k, so \frac{dk}{dT} = k\cdot\frac{E_a}{RT^{2}}, \qquad\text{i.e.}\qquad \frac{1}{k}\frac{dk}{dT} = \frac{E_a}{RT^{2}}. Numerically, \dfrac{E_a}{RT^{2}} = \dfrac{75\,000}{8.314 \times 298.15^{2}} = 0.101 K^{-1}: the rate constant grows by about 10 % per kelvin.
Check. Units: E_a/(RT^{2}) is (J/mol)/((J/mol·K)·K²) = K^{-1}, a per-kelvin fractional rate — consistent. Magnitude: a 10 % gain per kelvin means the rate doubles about every \ln 2/0.101 \approx 6.8 K; for E_a = 55 kJ/mol the same calculation gives 9.3 K — the familiar “rate doubles every ten degrees” rule of thumb, now seen to belong to a particular activation energy rather than to chemistry in general. (Over a finite 10 K the exact ratio is 2.67, while the derivative extrapolated naively gives 2.76 — a local rate is a good guide, not an identity.)
Piece 3: the logarithm — the straight line explained
The logarithm’s derivative follows from the exponential by the chain rule. If y = \ln x then e^{y} = x; differentiating both sides with respect to x and applying the chain rule to the left gives e^{y}\,y' = 1, so y' = 1/e^{y} = 1/x.
Theorem 4.2.4. \;\dfrac{d}{dx}\ln x = \dfrac{1}{x} for x > 0, and by the chain rule \dfrac{d}{dt}\ln f(t) = \dfrac{f'(t)}{f(t)}.
The second form says that the derivative of \ln f is the fractional rate of change of f — the rate divided by the amount. Apply it to first-order decay: \frac{d}{dt}\ln[\mathrm{A}] = \frac{d[\mathrm{A}]/dt}{[\mathrm{A}]} = \frac{-k[\mathrm{A}]}{[\mathrm{A}]} = -k. A constant. A function whose derivative is constant everywhere is a straight line, so \ln[\mathrm{A}] plotted against t is exactly straight with slope -k. This is the proof of the diagnostic you used on the exam: the ln-plot’s slope is -k because the fractional rate of a first-order reaction does not change. A curved ln-plot means the fractional rate is changing — a different order, or a different mechanism.
Gluing 3: the product rule — a rectangle that grows
When two quantities that both change are multiplied — the encounter factor q = [\mathrm{A}][\mathrm{B}], say, when both concentrations are changing — the product’s rate has two contributions. Picture a rectangle of area A\times B. Let the width grow by \Delta A and the height by \Delta B:
The new area is the old rectangle plus two thin strips plus a tiny corner: \Delta(AB) = B\,\Delta A + A\,\Delta B + \Delta A\,\Delta B. Divide by \Delta t and shrink. The two strips each keep a factor of one small quantity over \Delta t and survive; the corner carries two small factors, one of which is left over after dividing, so it vanishes.
Theorem 4.2.5 (product rule). \;\dfrac{d}{dx}\big[f(x)\,g(x)\big] = f'(x)\,g(x) + f(x)\,g'(x).
Two chemical uses are worth fixing now. First, if A is consumed at rate -0.20 M/min while B is consumed at -0.05 M/min, with [\mathrm{A}]=1.0 M and [\mathrm{B}] = 0.60 M, the encounter factor changes at (-0.20)(0.60) + (1.0)(-0.05) = -0.17 M²/min — the sum of “A’s change times B’s level” and “B’s change times A’s level.” Second, and with a view to the next but one section: if the two concentrations are tied by a fixed total, [\mathrm{B}] = C - [\mathrm{A}], then \dfrac{d}{d[\mathrm{A}]}\big([\mathrm{A}](C-[\mathrm{A}])\big) = (1)(C - [\mathrm{A}]) + [\mathrm{A}](-1) = C - 2[\mathrm{A}], which vanishes at an even split. That one line is the whole of §4.4.
Division needs no new rule: f/g = f\cdot g^{-1}, so the product rule, the power rule, and the chain rule together give \frac{d}{dx}\frac{f}{g} = \frac{f'g - fg'}{g^{2}}.
Example 4.2.2 — diminishing returns in an enzyme. The Michaelis–Menten rate is v(S) = \dfrac{V_{\max}\,S}{K_m + S}. With V_{\max} = 10\ \mu\text{M/s} and K_m = 2.0 mM, how much does the rate gain per extra millimolar of substrate at S = 0.5, 2, and 8 mM?
Setting it up. Write v = V_{\max}\,S\,(K_m + S)^{-1}: a product of S with a nested power, so product rule, power rule, and chain rule all appear.
Solution. \frac{dv}{dS} = V_{\max}\left[(K_m+S)^{-1} - S\,(K_m+S)^{-2}\right] = \frac{V_{\max}\,K_m}{(K_m+S)^{2}}. Numerically (in \muM s^{-1} mM^{-1}): at S = 0.5: 3.20; at S = 2: 1.25; at S = 8: 0.20.
Check. Units: (μM/s) per mM — rate per concentration, as a derivative should be. Limiting cases: dv/dS is always positive (more substrate never slows the enzyme) and falls toward zero as S \to \infty — the saturation plateau. At S = K_m = 2 mM the rate is exactly half of V_{\max} (v = 5.00\ \muM/s), which is the definition of K_m. And the marginal benefit of substrate fell by a factor of sixteen between 0.5 and 8 mM while the rate itself only rose from 2.0 to 8.0: diminishing returns, quantified.
Remarks and cautions
The chain factor. \dfrac{d}{dt}\,0.80\,e^{-0.139t} is not 0.80\,e^{-0.139t}. The exponential reproduces itself only when the exponent is the bare variable; otherwise the derivative of the exponent multiplies. Two instant tests catch the omission: the sign (a falling concentration cannot have a positive derivative) and the units (the result must carry “per minute”, and only the factor -0.139 min^{-1} can supply it).
The product rule does not act term by term. (fg)' \neq f'g'. Test it on f=g=x: (x^{2})' = 2x, but f'g' = 1. The product rule has two terms because both factors are changing — two strips of the rectangle.
Power or exponential? In x^{n} the variable is the base; in e^{ax} it is the exponent. \dfrac{d}{dx}\,e^{x} \neq x\,e^{x-1}. A concentration that depends on time through an exponent is an exponential, and its derivative stays proportional to itself.
Summary
Chemistry’s functions are assemblies of powers, exponentials, and logarithms combined by addition, scaling, multiplication, and nesting, so a handful of derivations buys the derivative of nearly anything we will meet. Sums and constants pass through (rates add); powers lose one degree and gain the old degree as a coefficient, which is how P \propto V^{-1} gives the pressure’s compression sensitivity and a 1/r energy gives an inverse-square force; the exponential differentiates to a multiple of itself; and a nested function differentiates by multiplying the rates of the two links in the chain — a rule whose units alone expose the commonest mistake of dropping the inner factor. From the chain rule come the logarithm’s derivative 1/x and the proof that a first-order ln-plot is a straight line of slope -k; from the rectangle comes the product rule and, through it, the quotient rule behind Michaelis–Menten saturation. Rules like these can find where a function turns, and the next section does exactly that. What they cannot yet handle is a turning point with a budget attached — two quantities that must share a fixed total — which is the subject of §4.4.
Practice until it sticks
Problem Set 6 applies these rules to the chemistry and physics functions of this section; Problems 7 and 10 target the chain factor directly.