Steady State
MAT161 / Lecture 6
Lecture 6 · Barnett ch. 2 · about 40 min

The Derivative: Slopes, Rates and Marginal Cost

What a derivative really measures, how to compute one from its definition and with the basic rules, and why economists call it the marginal cost or marginal revenue.

By the end you can
  • Compute an average rate of change as the slope of a secant line
  • Explain the derivative as the limit of secant slopes, and compute it from the definition
  • Differentiate polynomials with the constant, power, constant-multiple and sum rules
  • Use marginal cost, revenue and profit to estimate the effect of one more unit

How fast is it changing?

A firm’s cost rises from 5,000 to 5,900 lira when output goes from 100 to 110 units. On average, each extra unit cost 90 lira. That is an average rate of change: the change in the output of a function divided by the change in its input.

Average rate of changeFormula sheet →
f(a+h)−f(a)h\frac{f(a + h) - f(a)}{h}
aa
starting input
hh
change in the input
f(a+h)−f(a)f(a+h) - f(a)
change in the output

Use it when a question asks for an average rate over an interval, such as average cost per extra unit between two output levels.

Geometrically, this is the slope of the straight line through the two points (a,f(a))(a, f(a)) and (a+h,f(a+h))(a+h, f(a+h)). That line is called a secant line.

But a manager deciding whether to produce one more unit does not care about the average over the last ten. They want the rate of change right now, at the current output. That is what a derivative measures.

From secant to tangent

Make hh smaller and the second point slides toward the first. The secant line swings, and its slope settles on a single number: the slope of the tangent line, the line that just touches the curve at that point.

Try it. Start with f(x)=x2f(x) = x^2 at a=1.5a = 1.5, then shrink hh and read the table.

A few things to notice:

  1. For x2x^2 at a=1.5a = 1.5, the table approaches 3, which is 2×1.52 \times 1.5. Move PP and check that the limit is always 2a2a.
  2. Switch to the cost function. The tangent slope at output xx is x+1x + 1: the cost of the next unit.
  3. On x3−3xx^3 - 3x, put PP at a peak or a valley. The tangent is flat there, so the derivative is zero. You will use this for optimisation later in the course.
Definition of the derivativeDefinitionFormula sheet →
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
f′(x)f'(x)
derivative of f at x, the instantaneous rate of change
h→0h \to 0
the step shrinks toward zero but never equals it

Use it when a question says from the definition or using the limit. Otherwise use the rules below, which are much faster.

The derivative has three equivalent readings, and exams use all three words:

Check your understanding

On the graph of ff, the secant slope from x=2x = 2 to x=2+hx = 2 + h is 4+h4 + h for every h≠0h \neq 0. What is f′(2)f'(2)?

Computing a derivative from the definition

The definition always follows the same four steps. Learn the routine and exam questions of this type become mechanical.

Worked example · The four-step process0/5

Find f′(x)f'(x) for f(x)=x2+3xf(x) = x^2 + 3x from the definition.

Your turn

Use the four-step process on f(x)=3x2−5f(x) = 3x^2 - 5. What is f′(2)f'(2)?

The basic rules

Nobody computes derivatives from the definition every time. Four rules cover every polynomial.

Constant ruleFormula sheet →
ddx k=0\frac{d}{dx}\,k = 0
kk
any constant

Use it when a term has no x. A flat line has slope zero, so fixed costs vanish from marginal cost.

Power ruleMost usedFormula sheet →
ddx xn=n xn−1\frac{d}{dx}\,x^n = n\,x^{n-1}
nn
any real number, including negatives and fractions

Use it when you see x to a power. Rewrite roots and fractions first: the square root of x is x to the 1/2, and 1/x is x to the -1.

Constant multiple and sum rulesFormula sheet →
ddx[k f(x)]=k f′(x)ddx[f(x)±g(x)]=f′(x)±g′(x)\frac{d}{dx}\big[k\,f(x)\big] = k\,f'(x) \qquad \frac{d}{dx}\big[f(x) \pm g(x)\big] = f'(x) \pm g'(x)
kk
constant
f,gf, g
differentiable functions

Use it when always. Differentiate a polynomial term by term and keep the coefficients.

Worked example · Rewrite first, then differentiate0/6

Differentiate f(x)=4x3−2x+6x−7f(x) = 4x^3 - \frac{2}{x} + 6\sqrt{x} - 7.

Check your understanding

What is the derivative of f(x)=1x3f(x) = \frac{1}{x^3}?

Your turn

f(x)=2x4−3x2+8x−1f(x) = 2x^4 - 3x^2 + 8x - 1. What is f′(1)f'(1)?

Marginal analysis

In economics the derivative gets a name of its own. If C(x)C(x) is the total cost of producing xx units, then C′(x)C'(x) is the marginal cost. It approximates the cost of producing one more unit, the (x+1)(x+1)-th.

Marginal cost, revenue and profitEconomicsFormula sheet →
C′(x)≈C(x+1)−C(x)P(x)=R(x)−C(x)  ⇒  P′(x)=R′(x)−C′(x)C'(x) \approx C(x+1) - C(x) \qquad P(x) = R(x) - C(x) \;\Rightarrow\; P'(x) = R'(x) - C'(x)
C(x)C(x)
total cost
R(x)R(x)
total revenue
P(x)P(x)
total profit
xx
units produced

Use it when a question asks for the approximate cost, revenue or profit of the next unit. Use the exact difference only if it asks for the exact change.

Why only approximately? The marginal cost is the slope of the tangent at xx. The true cost of the next unit is the slope of the secant from xx to x+1x + 1: an h=1h = 1 secant. When the curve bends slowly, the two are very close.

Worked example · Marginal versus exact cost0/5

A company's cost of producing xx units is C(x)=10,000+150x−0.2x2C(x) = 10{,}000 + 150x - 0.2x^2. Find the marginal cost at x=200x = 200 and compare it with the exact cost of the 201st unit.

Your turn

Revenue from selling xx units is R(x)=60x−0.02x2R(x) = 60x - 0.02x^2. What is the marginal revenue at x=1,000x = 1{,}000?

When there is no derivative

A derivative is a limit, and limits can fail to exist. At three kinds of points f′(a)f'(a) does not exist:

Check your understanding

Which statement is always true?

Exam practice

Exam question 1

f(x)=x2−4xf(x) = x^2 - 4x. Find the average rate of change of ff from x=1x = 1 to x=4x = 4.

Exam question 2

Find the slope of the tangent line to y=x+8xy = \sqrt{x} + \frac{8}{x} at x=4x = 4.

Exam question 3

C(x)=800+12x+0.05x2C(x) = 800 + 12x + 0.05x^2 and R(x)=40xR(x) = 40x. Find the marginal profit at x=100x = 100.

Exam question 4

For a cost function, C′(500)=8C'(500) = 8. Which interpretation is correct?

Summary and review

Review deck · 14 cards0/14 mastered