Limits marginal change and optimization

The cost of producing one more item is a finite difference. A derivative instead describes the limiting cost per unit of a very small change. The distinction matters when a model uses continuous quantities to approximate discrete decisions. We will use calculus to predict changes, then check the candidate decision against the feasible domain.

1 Limits continuity and derivatives

For a function defined near $a$ except possibly at $a$, $\lim_{x\to a}f(x)=L$ means: for every $\varepsilon>0$ there is $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-L|<\varepsilon$. A function is continuous at $a$ if this limit equals its defined value $f(a)$.

The derivative is

$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h,$$

when the finite limit exists. Thus $f(a+h)=f(a)+f'(a)h+o(|h|)$: the error divided by $|h|$ tends to zero. Differentiability implies continuity, since the increment is $h$ times a bounded difference quotient near zero. Continuity does not imply differentiability: $|x|$ has one-sided slopes $-1$ and $1$ at zero.

For $f(x)=x^2$, the quotient is $2a+h$, so $f'(a)=2a$. Limit laws give the sum and scalar rules. Expanding an increment of $fg$ gives the product rule $(fg)'=f'g+fg'$. Composing local linear approximations gives the chain rule $(f\circ g)'(a)=f'(g(a))g'(a)$ whenever both derivatives exist. The quotient rule requires a nonzero denominator.

Example. If $C(q)=100+4q+0.02q^2$, then $C'(q)=4+0.04q$. At $q=50$, marginal cost is 6 currency units per item. The exact next-item cost is $C(51)-C(50)=6.02$, not 6. The linear estimate omits the quadratic remainder $0.02(1)^2$.

2 Mean value and what a derivative tells us

Mean value theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $a<b$, there is $c\in(a,b)$ with

$$f(b)-f(a)=f'(c)(b-a).$$

Proof. Subtract the secant line: $g(x)=f(x)-f(a)-[f(b)-f(a)](x-a)/(b-a)$. Then $g(a)=g(b)=0$. By the extreme-value theorem, either $g$ is constant or a nonzero maximum or minimum occurs inside the interval. At an interior differentiable extremum, positive and negative difference quotients force $g'(c)=0$. Rearranging gives the result. $\square$

Consequently, $f'\ge0$ throughout an interval implies $f$ is nondecreasing there. If $|f'|\le M$, then $|f(b)-f(a)|\le M|b-a|$. A derivative evaluated at one point alone does not provide such an interval-wide bound.

3 Profit and constrained decisions

Let demand price be $p(q)=30-0.1q$ and cost $C(q)=100+6q+0.05q^2$, with a model domain $0\le q\le100$. Revenue is $R(q)=q\,p(q)$, not merely $p(q)$. Profit is

$$\Pi(q)=R(q)-C(q)=-100+24q-0.15q^2.$$

Its derivative vanishes at $q=80$. Since $\Pi''=-0.3<0$, the derivative is positive before 80 and negative after 80, so 80 is the unique global maximizer on the domain. Its profit is 860, compared with $-100$ at zero and 800 at 100. This is an optimal decision in the invented model, not a recommendation about a real business.

An interior differentiable optimum must have derivative zero. The converse is false: $x^3$ has derivative zero at zero without a local extremum. Endpoints and nondifferentiable feasible points must also be checked. For integer quantities one compares nearby feasible integers and any relevant boundaries rather than announcing a fractional production plan.

For positive $f(q)$ and $q>0$, the dimensionless elasticity is $E(q)=qf'(q)/f(q)$. It approximates relative response: $\Delta f/f\approx E\,\Delta q/q$. It is undefined when $f(q)=0$ and is not an exact finite-percentage formula.

4 Graded practice

E1 Foundation. Derive the derivative of $1/x$ at a nonzero $a$ directly from the definition.

E2 Marginal interpretation. For $C(q)=40+5q+q^2/10$, compare $C'(10)$ with $C(11)-C(10)$.

E3 Optimization. Maximize $P(q)=-q^2+14q-10$ on $[0,5]$, then on $[0,10]$.

E4 Proof. If $f'(x)=0$ on an interval, prove $f$ is constant there. Explain why the conclusion need not hold on a disconnected domain.

5 Full solutions

S1. For sufficiently small $h$ with $a+h\ne0$, the difference quotient is $[1/(a+h)-1/a]/h=-1/[a(a+h)]$, tending to $-1/a^2$.

S2. $C'(q)=5+q/5$, so the derivative is 7. The exact increment is $5+(121-100)/10=7.1$. Units are currency per unit for the derivative and currency for the one-unit increment; their numerical comparison uses an increment of one unit.

S3. $P'=14-2q$. It is positive throughout $[0,5)$, so the first maximum is $P(5)=35$. On $[0,10]$ it changes from positive to negative at 7, giving maximum $P(7)=39$. The stationary point outside the first feasible interval cannot be chosen there.

S4. For any two points in the interval, the mean value theorem gives $f(b)-f(a)=f'(c)(b-a)=0$. On $(-2,-1)\cup(1,2)$, the function equal to 0 on the first component and 1 on the second has derivative zero everywhere on its domain but is not constant on that domain.

Summary

A derivative is a local linear rate; a finite difference is an actual change. The mean value theorem links rates to interval-wide behavior. Optimization requires the model domain, boundary candidates and an argument that the chosen stationary point really maximizes the objective.