4 Derivative-Based Analysis

A progression from the meaning and computation of derivatives to function analysis, marginal quantities, optimization, and responsible interpretation of quantitative results.

Rates of Change and Derivatives

A measures how rapidly an output changes as its input changes. For a function y=f(x)y=f(x), the at an input is

f′(x)=lim⁡h→0f(x+h)−f(x)h.f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.

This differs from the average rate of change over [a,b][a,b]:

f(b)−f(a)b−a.\frac{f(b)-f(a)}{b-a}.

The average rate describes change across an interval, whereas the describes behavior at one input. Geometrically, f′(a)f'(a) is the slope of the tangent line to the graph at x=ax=a.

Units provide an important interpretation check. A has units of output divided by input. If s(t)s(t) is position in meters and tt is time in seconds, then s′(t)s'(t) has units of meters per second and represents velocity, while s′′(t)s''(t) has units of meters per second squared and represents acceleration.

A positive indicates instantaneous growth, and a negative indicates instantaneous decline. The magnitude describes how rapidly the change occurs, but its meaning depends on the quantities and units in the model.

Takeaway: A is both a geometric slope and an instantaneous rate with units.

Differentiation Rules

Differentiation rules turn the limit definition into an efficient calculation method. For a constant CC,

ddx(C)=0.\frac{d}{dx}(C)=0.

For any real number nn, the power rule is

ddx(xn)=nxn−1.\frac{d}{dx}(x^n)=nx^{n-1}.

The constant multiple, sum, and difference rules are

ddx[Cf(x)]=Cf′(x),\frac{d}{dx}[Cf(x)]=Cf'(x),
ddx[f(x)+g(x)]=f′(x)+g′(x),\frac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x),
ddx[f(x)−g(x)]=f′(x)−g′(x).\frac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x).

For a product, apply the :

ddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x).\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x).

For a quotient, apply the quotient rule:

ddx[f(x)g(x)]=f′(x)g(x)−f(x)g′(x)[g(x)]2,g(x)≠0.\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2},\qquad g(x)\ne 0.

For a composite function, apply the . If y=f(g(x))y=f(g(x)), then

ddxf(g(x))=f′(g(x))g′(x).\frac{d}{dx}f(g(x))=f'(g(x))g'(x).

For example,

ddx(3x2+1)4=4(3x2+1)3(6x)=24x(3x2+1)3.\frac{d}{dx}(3x^2+1)^4=4(3x^2+1)^3(6x)=24x(3x^2+1)^3.

When several rules apply, identify the structure first: sums use the sum rule, products use the , quotients use the quotient rule, and nested expressions use the .

Takeaway: Accurate differentiation begins with recognizing the algebraic structure before selecting a rule.

First- Analysis

The sign of the first organizes a function's behavior over intervals. If f′(x)>0f'(x)>0 throughout an interval, ff is increasing there. If f′(x)<0f'(x)<0, ff is decreasing. If f′(x)=0f'(x)=0 throughout an interval, the function is constant.

Consider

f(x)=x3−3x2+2.f(x)=x^3-3x^2+2.

Its is

f′(x)=3x2−6x=3x(x−2).f'(x)=3x^2-6x=3x(x-2).

The critical numbers are x=0x=0 and x=2x=2. Testing the sign of f′f' on the resulting intervals gives:

  • On (−∞,0)(- \infty,0), f′(x)>0f'(x)>0, so the function is increasing.

  • On (0,2)(0,2), f′(x)<0f'(x)<0, so the function is decreasing.

  • On (2,∞)(2,\infty), f′(x)>0f'(x)>0, so the function is increasing.

A is an input in the domain where the is zero or undefined. It is a candidate for a change in behavior, not automatically an extremum.

The first test classifies a by its sign change:

  • Positive to negative: .

  • Negative to positive: local minimum.

  • No sign change: neither a nor a local minimum.

For this example, the change from positive to negative at x=0x=0 gives a with value f(0)=2f(0)=2. The change from negative to positive at x=2x=2 gives a local minimum with value f(2)=−2f(2)=-2. In contrast, f(x)=x3f(x)=x^3 has f′(x)=3x2f'(x)=3x^2, which is nonnegative on both sides of x=0x=0; the does not produce a local extremum.

Takeaway: Use a sign chart for f′f' to identify increasing and decreasing intervals and to classify critical points.

and the Second

The second tracks how the first changes:

f′′(x)=ddxf′(x).f''(x)=\frac{d}{dx}f'(x).

If f′′(x)>0f''(x)>0, slopes are increasing and the graph is concave up. If f′′(x)<0f''(x)<0, slopes are decreasing and the graph is concave down.

A possible inflection point occurs where f′′(x)=0f''(x)=0 or where f′′(x)f''(x) is undefined. It is an actual inflection point only if the changes there.

For

f(x)=x3−3x2+2,f(x)=x^3-3x^2+2,

we have

f′′(x)=6x−6.f''(x)=6x-6.

Therefore, the graph is concave down for x<1x<1 and concave up for x>1x>1. Because the changes at x=1x=1, the graph has an inflection point there.

The second also provides a quick test for a critical point where f′(c)=0f'(c)=0:

  • If f′′(c)>0f''(c)>0, the function has a local minimum at x=cx=c.

  • If f′′(c)<0f''(c)<0, the function has a at x=cx=c.

  • If f′′(c)=0f''(c)=0, the test is inconclusive; use the first test.

and increasing or decreasing behavior answer different questions. A function can be increasing while concave down if its positive slope is becoming smaller, or decreasing while concave up if its negative slope is becoming less negative.

Takeaway: The first describes the direction of change; the second describes how that direction is changing.

Marginal Analysis and Optimization

Derivatives support local approximations and optimization, but conclusions must respect the domain and the model's assumptions.

If C(q)C(q) is the cost of producing qq units, then C′(q)C'(q) is the . For a small production change Δq\Delta q,

ΔC≈C′(q)Δq.\Delta C\approx C'(q)\Delta q.

Suppose

C(q)=500+12q+0.02q2.C(q)=500+12q+0.02q^2.

Then

C′(q)=12+0.04q.C'(q)=12+0.04q.

At q=100q=100,

C′(100)=16.C'(100)=16.

The model therefore estimates an additional cost of approximately 1616 dollars for one more unit near a production level of 100100 units. This is a local approximation, not necessarily the exact cost of the next unit. The same reasoning applies to marginal revenue and marginal profit.

For global optimization on a closed interval, use a complete comparison procedure:

  1. Find all critical numbers in the interior of the interval.

  2. Evaluate the function at each .

  3. Evaluate the function at both endpoints.

  4. Compare all values; the largest is the and the smallest is the global minimum.

For a rectangular garden with perimeter 4040 meters, let one side be xx. The other side is 20−x20-x, so

A(x)=x(20−x)=20x−x2,0≤x≤20.A(x)=x(20-x)=20x-x^2,\qquad 0\le x\le 20.

Differentiating gives

A′(x)=20−2x.A'(x)=20-2x.

The interior critical point satisfies

20−2x=0⇒x=10.20-2x=0\quad\Rightarrow\quad x=10.

Since A′′(x)=−2<0A''(x)=-2<0, the area is maximized at x=10x=10. The dimensions are 1010 meters by 1010 meters, and the maximum area is 100100 square meters. Checking the endpoints confirms the global result on the stated domain.

A responsible interpretation should identify the measured quantity, state units, give the sign and magnitude, specify the relevant point or interval, distinguish exact results from approximations, and acknowledge that the conclusion describes the model rather than automatically proving that the model represents reality.

Takeaway: Optimization requires critical-point analysis, endpoint checks, domain awareness, and a clear interpretation of units and approximations.