2.1 Introduction to Differentiation

While functions allow us to model the state of a system—such as profit at a given price—differentiation provides the mathematical tools to analyze its dynamics, or rates of change. In business and economics, understanding rates of change is critical. We are not just interested in the total cost, but in the marginal cost—the change in cost from producing one more unit. We don’t just want to know the profit, but the price at which profit is maximized. This chapter introduces the concept of the derivative, which is the mathematical formalization of an instantaneous rate of change, and explores its powerful applications.

2.2 The Derivative: A Formal Definition

Defining the Derivative

For a function y = f(x), the derivative of f at x, denoted f'(x), is the instantaneous rate of change of the function at that point. It is formally defined as the limit of the average rate of change over an infinitesimally small interval:

f'(x) = lim Δx→0 [f(x + Δx) – f(x)] / Δx

This limit, if it exists, tells us the precise slope of the line tangent to the function’s graph at the point x.

The Calculation Process

To find a derivative using the formal definition, we follow a three-step process:

1. Form the difference quotient: [f(x + Δx) – f(x)] / Δx.
2. Simplify the difference quotient algebraically.
3. Calculate the limit of the simplified expression as Δx approaches 0.

Illustrative Examples

Example 1: Finding the Derivative of f(x) = x²

• Step 1: Form the difference quotient. [f(x + Δx) – f(x)] / Δx = [(x + Δx)² – x²] / Δx
• Step 2: Simplify. = [x² + 2x(Δx) + (Δx)² – x²] / Δx = [2x(Δx) + (Δx)²] / Δx = 2x + Δx
• Step 3: Calculate the limit. f'(x) = lim Δx→0 (2x + Δx) = 2x + 0 = 2x Thus, the derivative of f(x) = x² is f'(x) = 2x.

Example 2: Profit Maximization A manufacturer’s profit from selling radios at price x is P(x) = -400x² + 6800x – 12000. To find the price that maximizes profit, we must find where the peak of this profit parabola occurs. This is the point where the slope of the tangent line is zero—that is, where the derivative is zero.

• Step 1 & 2: Calculate the derivative P'(x) using the limit definition. (The full algebraic expansion is lengthy, but it follows the same pattern as the x² example.) The simplified result is: P'(x) = lim Δx→0 [-800x – 400(Δx) + 6800]
• Step 3: Find the limit. P'(x) = -800x – 400(0) + 6800 = -800x + 6800
• Step 4: Find the price that maximizes profit. Set the derivative to zero and solve for x. P'(x) = 0 -800x + 6800 = 0 6800 = 800x x = 6800 / 800 = 8.5 The optimal selling price is $8.50 per radio. Note that this is the calculus-based method for finding the vertex of the profit parabola, a concept we previously approached using the algebraic formula x = -b/2a. Both methods yield the peak of the parabola, but the derivative is a tool that can be applied to functions of any shape.

2.3 Techniques of Differentiation

While the limit definition is the theoretical foundation of the derivative, applying it every time is tedious. A set of rules derived from this definition makes the process of finding derivatives much more efficient.

The Differentiation Rules

• The Power Rule: d/dx(xⁿ) = nxⁿ⁻¹
  • Explanation: To find the derivative of x raised to a power, bring the power down as a multiplier and then reduce the original power by one.
  • Example 1:
    • a. d/dx(x²⁷) = 27x²⁶
    • b. d/dx(1/x²⁷) = d/dx(x⁻²⁷) = -27x⁻²⁸
    • c. d/dx(√x) = d/dx(x¹/²) = (1/2)x⁻¹/² = 1 / (2√x)
    • d. d/dx(1/√x) = d/dx(x⁻¹/²) = (-1/2)x⁻³/²
• The Derivative of a Constant: d/dx(C) = 0
  • Explanation: A constant value does not change, so its rate of change is always zero.
• The Constant Multiple Rule: d/dx[C * f(x)] = C * f'(x)
  • Explanation: The derivative of a constant times a function is the constant times the derivative of the function.
  • Example 2: d/dx(3x⁵) = 3 * d/dx(x⁵) = 3 * (5x⁴) = 15x⁴
• The Sum Rule: d/dx[f(x) ± g(x)] = f'(x) ± g'(x)
  • Explanation: The derivative of a sum or difference of functions is the sum or difference of their individual derivatives.
  • Example 3: d/dx(x² + 3x⁵) = d/dx(x²) + d/dx(3x⁵) = 2x + 15x⁴
• The Product Rule: d/dx[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)
  • Explanation: The derivative of a product is the first function times the derivative of the second, plus the second function times the derivative of the first.
  • Example 4: Differentiate y = (x²)(3x + 1) y’ = (x²) * d/dx(3x + 1) + (3x + 1) * d/dx(x²) y’ = (x²)(3) + (3x + 1)(2x) = 3x² + 6x² + 2x = 9x² + 2x
• The Quotient Rule: d/dx[f(x)/g(x)] = [g(x)f'(x) – f(x)g'(x)] / [g(x)]²
  • Explanation: The derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator. A common mnemonic to remember this is: “Low d-high minus high d-low, square the bottom and away we go.”
  • Example 5: Differentiate y = (x²+2x-21) / (x-3) y’ = [(x-3) * d/dx(x²+2x-21) – (x²+2x-21) * d/dx(x-3)] / (x-3)² y’ = [(x-3)(2x+2) – (x²+2x-21)(1)] / (x-3)² y’ = [2x²-4x-6 – x²-2x+21] / (x-3)² = (x²-6x+15) / (x-3)²

These rules provide the mechanical tools for differentiation; now we return to the conceptual meaning of the derivative as a rate of change.

2.4 The Derivative as a Rate of Change

The derivative provides a precise, mathematical measure of instantaneous rate of change, a concept that is distinct from the average rate of change over an interval.

Average vs. Instantaneous Rate of Change

• The average rate of change of a function f(x) over an interval from x to x + Δx is given by the difference quotient: Average Rate = Δy / Δx = [f(x + Δx) – f(x)] / Δx
• The instantaneous rate of change is the limit of this average rate as the interval Δx shrinks to zero. This is precisely the definition of the derivative: Instantaneous Rate = dy/dx = f'(x)

Population Example Analysis

Example 1: A community’s population x months from now is estimated to be P(x) = x² + 20x + 8,000.

• (a) At what rate will the population be changing 15 months from now? We need the instantaneous rate of change at x=15. First, find the derivative: P'(x) = 2x + 20 Now, evaluate it at x=15: P'(15) = 2(15) + 20 = 30 + 20 = 50 Interpretation: 15 months from now, the population will be growing at a rate of 50 people per month.
• (b) By how much will the population actually change during the 16th month? The 16th month is the interval between x=15 and x=16. The actual change is: Actual Change = P(16) – P(15) P(16) = (16)² + 20(16) + 8000 = 256 + 320 + 8000 = 8576 P(15) = (15)² + 20(15) + 8000 = 225 + 300 + 8000 = 8525 Actual Change = 8576 – 8525 = 51 The actual change of 51 people is very close to the instantaneous rate of 50, illustrating how the derivative provides an excellent approximation.

Percentage Rate of Change

To provide a relative measure of change, we use the percentage rate of change, which compares the rate of change to the size of the quantity itself.

Percentage rate of change = 100 * [f'(x) / f(x)]

GNP Example Analysis

Example 2: A country’s Gross National Product (GNP) t years after 1980 was N(t) = t² + 5t + 106 billion dollars.

• (a) At what rate was the GNP changing in 1988? 1988 corresponds to t=8. The rate of change is the derivative N'(t) = 2t + 5. N'(8) = 2(8) + 5 = 21 Interpretation: In 1988, the GNP was growing at a rate of $21 billion per year.
• (b) At what percentage rate was the GNP changing in 1988? We also need the value of the GNP in 1988: N(8) = (8)² + 5(8) + 106 = 64 + 40 + 106 = 210. Percentage Rate = 100 * [N'(8) / N(8)] = 100 * [21 / 210] = 10 Interpretation: In 1988, the GNP was growing at a rate of 10 percent per year.

This ability of the derivative to provide a close approximation for change is formalized in the concepts of differentials and marginal analysis.

2.5 Approximation by Differentials and Marginal Analysis

Introduction to Approximation

The derivative can be used to estimate the change in a function, Δy, that results from a small change in its variable, Δx. This is one of the most powerful practical applications of calculus. The core idea is that for a small enough Δx, the change along the tangent line is a very good approximation of the change along the curve itself.

Approximation with Differentials

The approximation formula is: Δy ≈ (dy/dx) * Δx

Example 1: Manufacturing Cost The cost to manufacture q units is C(q) = 3q² + 5q + 10. If production is currently 40 units, estimate the change in cost if production increases to 40.5 units.

• Here, q=40 and Δq = 0.5.
• The derivative (rate of change) is C'(q) = 6q + 5.
• At the current production level, the rate is C'(40) = 6(40) + 5 = 245.
• The estimated change in cost is: ΔC ≈ C'(40) * Δq = 245 * 0.5 = $122.50 Interpretation: If production is increased from 40 to 40.5 units, the total cost will increase by approximately $122.50.

Example 2: Factory Output Daily output is Q(L) = 900L¹/³, where L is labor in worker-hours. Currently, 1,000 worker-hours are used. Estimate the additional labor needed to increase output by 15 units.

• Here, we know the desired change in output ΔQ = 15. We need to find the required change in input, ΔL.
• The formula is ΔQ ≈ Q'(L) * ΔL.
• The derivative is Q'(L) = 900 * (1/3)L⁻²/³ = 300L⁻²/³.
• At L=1000, the rate is Q'(1000) = 300(1000)⁻²/³ = 300(10⁻²) = 3.
• Rearranging the formula: ΔL ≈ ΔQ / Q'(L) = 15 / 3 = 5. Interpretation: Approximately 5 additional worker-hours are needed.

Approximation of Percentage Change

The formula for approximating percentage change is: Percentage change in f ≈ 100 * [f'(x) * Δx] / f(x)

• Example 3 (GNP): For N(t) = t² + 5t + 200, estimate the percentage change during the first quarter of 1998 (t=8, Δt=0.25).
  • % Change ≈ 100 * [N'(8) * 0.25] / N(8) = 100 * [(2*8+5) * 0.25] / (8²+5*8+200) = 100 * [21 * 0.25] / 304 ≈ 1.73% (Note: Source text has an error, using 264 instead of 304 in the denominator.)
• Example 4 (Factory Output): For Q(K) = 4000K¹/², estimate the percentage increase in output from a 1% increase in capital (ΔK = 0.01K).
  • % Change ≈ 100 * [Q'(K) * ΔK] / Q(K) = 100 * [2000K⁻¹/² * (0.01K)] / (4000K¹/²) = 100 * [20K¹/²] / (4000K¹/²) = 0.5%

Marginal Analysis in Economics

Marginal analysis is the specific use of the derivative to approximate the change in a function resulting from a 1-unit change in its variable.

• Marginal Cost C'(x): The derivative of the total cost function.
• Marginal Revenue R'(x): The derivative of the total revenue function.

Key Interpretation:

• C'(x) approximates the cost of producing the (x+1)st unit.
• R'(x) approximates the revenue from selling the (x+1)st unit.

Example 5 Analysis Cost is C(x) = (1/8)x² + 3x + 98. Price is P(x) = (1/3)(75-x).

• (a) Find Marginal Cost and Revenue:
  • Marginal Cost: C'(x) = 2(1/8)x + 3 = (1/4)x + 3.
  • Total Revenue: R(x) = x * P(x) = x(1/3)(75-x) = 25x – (1/3)x².
  • Marginal Revenue: R'(x) = 25 – (2/3)x.
• (b) Estimate the cost of producing the 9th unit. This is approximated by the marginal cost at x=8. C'(8) = (1/4)(8) + 3 = 2 + 3 = $5.
• (c) What is the actual cost of producing the 9th unit? The actual cost is C(9) – C(8). C(9) = (1/8)(9)² + 3(9) + 98 = 10.125 + 27 + 98 = 135.125. C(8) = (1/8)(8)² + 3(8) + 98 = 8 + 24 + 98 = 130. Actual Cost = 135.125 – 130 = $5.125. The estimate is quite close.
• (d) Estimate the revenue from the 9th unit. This is approximated by the marginal revenue at x=8. R'(8) = 25 – (2/3)(8) = 25 – 16/3 ≈ $19.67.
• (e) What is the actual revenue from the 9th unit? The actual revenue is R(9) – R(8). R(9) = 25(9) – (1/3)(9)² = 225 – 27 = 198. R(8) = 25(8) – (1/3)(8)² = 200 – 64/3 ≈ 178.67. Actual Revenue = 198 – 178.67 = $19.33.

When dealing with more complex functions, such as those where cost depends on output which in turn depends on time, we need an additional technique called the chain rule.

2.6 The Chain Rule

Introduction to the Chain Rule

The chain rule is essential for differentiating composite functions. Consider a scenario where the total cost C is a function of the number of units produced q, and q is itself a function of time t. To find the rate of change of cost with respect to time, dC/dt, we need a way to link these rates. The chain rule provides this link.

The Formula

If y is a function of u, and u is a function of x, then y is a composite function of x. The chain rule states: dy/dx = (dy/du) * (du/dx)

Illustrative Examples

Example 1 & 2: If y = u³ + u and u = 7x + 1, we find the derivative of each part:

• dy/du = 3u² + 1
• du/dx = 7 Applying the rule: dy/dx = (3u² + 1) * 7. We can then substitute u=7x+1 back in for a final answer in terms of x.

Example 3: Differentiate f(x) = (x² + 3x + 2)⁵.

• Let u = x² + 3x + 2, so f(u) = u⁵.
• df/du = 5u⁴ and du/dx = 2x + 3.
• f'(x) = (df/du) * (du/dx) = 5u⁴ * (2x + 3) = 5(x² + 3x + 2)⁴(2x + 3).

Example 4: Carbon Monoxide Revisited The carbon monoxide level is C(p) = √(0.5p + 17) parts per million, where p is population in thousands. The population t years from now is p(t) = 3.1 + 0.1t². Find the rate the carbon monoxide level is changing 3 years from now.

• Goal: Find dC/dt when t=3.
• Step 1: Find the individual derivatives.
  • dC/dp = (1/2)(0.5p + 17)⁻¹/² * (0.5) = 0.25 / √(0.5p + 17)
  • dp/dt = 0.2t
• Step 2: Apply the chain rule. dC/dt = (dC/dp) * (dp/dt) = [0.25 / √(0.5p + 17)] * (0.2t)
• Step 3: Evaluate at t=3. First, find the population at t=3: p(3) = 3.1 + 0.1(3)² = 3.1 + 0.9 = 4. Now, substitute t=3 and p=4 into the derivative: dC/dt = [0.25 / √(0.5*4 + 17)] * (0.2*3) = [0.25 / √19] * (0.6) ≈ 0.034 Interpretation: 3 years from now, the carbon monoxide level will be changing at a rate of approximately 0.034 parts per million per year. (Note: The final rate of ≈0.034 parts per million per year is derived directly from the stated functions. The source text arrives at a different numerical answer due to apparent errors in its own calculation.)

Having mastered the first derivative, we can now extend the concept to higher-order derivatives to gain even deeper insight into a function’s behavior.

2.7 Higher-Order Derivatives, Concavity, and Optimization

Introduction to Higher-Order Derivatives

Just as the first derivative measures the rate of change of a function, the second derivative measures the rate of change of the rate of change. In physics, if a function describes position, the first derivative is velocity and the second derivative is acceleration. In economics, the second derivative helps us understand if a growth rate is increasing or decreasing, a concept known as concavity.

The Second Derivative

The second derivative is found by differentiating the first derivative. It is denoted by f”(x) or d²y/dx².

Example 1: Calculation

• a. If f(x) = x³ – 12x + 1, then f'(x) = 3x² – 12, and f”(x) = 6x.
• b. If f(x) = 5x⁴ – 3x² – 3x + 7, then f'(x) = 20x³ – 6x – 3, and f”(x) = 60x² – 6.

Example 2: Worker Efficiency A worker’s output t hours after 8:00 AM is Q(t) = -t³ + 6t² + 24t.

• (a) Find the worker’s rate of production at 11:00 AM (t=3). Q'(t) = -3t² + 12t + 24 Q'(3) = -3(3)² + 12(3) + 24 = -27 + 36 + 24 = 33 units per hour.
• (b) At what rate is the rate of production changing at 11:00 AM? Q”(t) = -6t + 12 Q”(3) = -6(3) + 12 = -18 + 12 = -6 units per hour per hour. Interpretation: At 11:00 AM, the worker’s production rate is decreasing at a rate of 6 units per hour, every hour. The worker is getting tired.
• (c & d) Compare estimated vs. actual change. The estimated change in rate between 11:00 and 11:10 (1/6 of an hour) is ΔQ’ ≈ Q”(3) * Δt = -6 * (1/6) = -1 unit per hour. The actual change is Q'(3 + 1/6) – Q'(3) ≈ 31.92 – 33 = -1.08, showing a close approximation.

The nth Derivative

We can continue differentiating to find the third, fourth, or nth derivative, denoted f⁽ⁿ⁾(x). For y = 1/x = x⁻¹, the first five derivatives are y’=-x⁻², y”=2x⁻³, y”’=-6x⁻⁴, y⁽⁴⁾=24x⁻⁵, y⁽⁵⁾=-120x⁻⁶.

Concavity and Inflection Points

• Concavity: This describes the way a curve bends.
  • If f”(x) > 0, the graph is concave up (shaped like a cup holding water).
  • If f”(x) < 0, the graph is concave down (shaped like a cup spilling water).
• Critical Points: Points where f'(x) = 0 or is undefined. These are candidates for maxima or minima.
• Inflection Points: Points on the graph where the concavity changes (from up to down, or vice versa). These occur where f”(x) = 0 or is undefined.

Example 1: Finding Concavity and Inflection Points for f(x) = x³ – 3x² + 5x – 2

1. Find the second derivative: The source contains a typo. The correct first derivative is f'(x) = 3x² – 6x + 5. The second derivative is f”(x) = 6x – 6.
2. Find where f”(x) = 0: 6x – 6 = 0 gives x=1. This is our potential inflection point.
3. Test intervals:
   • For x < 1 (e.g., x=0), f”(0) = -6 < 0. The graph is concave down.
   • For x > 1 (e.g., x=2), f”(2) = 6 > 0. The graph is concave up.
4. Identify the inflection point: Since concavity changes at x=1, there is an inflection point there. The point is (1, f(1)) = (1, 1).

The Second-Derivative Test

This is a powerful tool for classifying critical points (f'(a)=0) as relative maxima or minima.

• If f”(a) > 0, the function is concave up at the critical point, so it must be a relative minimum.
• If f”(a) < 0, the function is concave down at the critical point, so it must be a relative maximum.
• If f”(a) = 0, the test is inconclusive.

Example 4: Point of Diminishing Returns A sales function is S(x) = -0.02x³ + 3x² + 100, where x is advertising spend. The “point of diminishing returns” is the inflection point where the graph changes from concave up (efficient spending) to concave down (inefficient spending).

• S'(x) = -0.06x² + 6x
• S”(x) = -0.12x + 6
• Set S”(x) = 0: -0.12x + 6 = 0 gives x = 50. This inflection point at x=50 (or $50,000) is the point of diminishing returns.

We now turn to a key application that synthesizes many of these derivative concepts: the elasticity of demand.

2.8 Key Applications: Elasticity and Revenue

This section explores one of the most important applications of derivatives in microeconomics: analyzing the sensitivity of demand to changes in price, a concept known as the elasticity of demand.

Elasticity of Demand

The elasticity of demand, denoted by the Greek letter eta (η), is a measure of how responsive the quantity demanded (q) is to a change in price (p). Conceptually, it represents the percentage change in demand generated by a 1% increase in price.

The formula for elasticity of demand is: η = (p/q) * (dq/dp)

Example 1 Analysis: Demand q and price p are related by q = 240 – 2p.

• (a) Express elasticity as a function of p. First, we need dq/dp, which is simply -2. η = (p / (240 – 2p)) * (-2) = -2p / (240 – 2p) = p / (p – 120)
• (b) Calculate elasticity when the price is p=100. η(100) = 100 / (100 – 120) = 100 / -20 = -5 Interpretation: When the price is $100, a 1% increase in price will cause an approximate 5% decrease in demand.
• (c) Calculate elasticity when the price is p=50. η(50) = 50 / (50 – 120) = 50 / -70 ≈ -0.71 Interpretation: When the price is $50, a 1% increase in price will cause an approximate 0.71% decrease in demand.
• (d) At what price is elasticity equal to -1? -1 = p / (p – 120) -(p – 120) = p -p + 120 = p 120 = 2p, so p = 60.

Levels of Elasticity

We classify elasticity based on its absolute value, |η|.

• Elastic (|η| > 1): Demand is highly sensitive to price changes. The percentage change in demand is greater than the percentage change in price.
• Inelastic (|η| < 1): Demand is not very sensitive to price changes. The percentage change in demand is less than the percentage change in price.
• Unit Elasticity (|η| = 1): The percentage change in demand is equal to the percentage change in price.

Elasticity and Total Revenue

The level of elasticity has a direct and critical relationship with total revenue (R = p*q).

• If demand is inelastic (|η| < 1), a price increase will increase total revenue. The drop in quantity sold is not enough to offset the higher price per unit.
• If demand is elastic (|η| > 1), a price increase will decrease total revenue. The drop in quantity sold is so large that it outweighs the higher price per unit.

Example 2 Analysis: Demand is given by q = 300 – p².

• (a) Determine where demand is elastic, inelastic, and unit elastic. First find η: dq/dp = -2p. So η = (p/q) * (dq/dp) = (p / (300 – p²)) * (-2p) = -2p² / (300 – p²).
  • Unit Elasticity: Set |η|=1. |-2p² / (300 – p²)| = 1 which gives 2p² = 300 – p², so 3p² = 300, p²=100, and p=10.
  • Inelastic: Demand is inelastic when |η|<1, which occurs for 0 ≤ p < 10.
  • Elastic: Demand is elastic when |η|>1, which occurs for 10 < p ≤ √300.
• (b) Describe the behavior of total revenue.
  • Revenue will increase as price increases on the inelastic interval (0, 10).
  • Revenue will decrease as price increases on the elastic interval (10, √300).
• (c) Confirm with calculus. The revenue function is R(p) = p*q = p(300 – p²) = 300p – p³. The derivative is R'(p) = 300 – 3p² = 3(100 – p²) = 3(10 – p)(10 + p). Setting R'(p) = 0 gives the critical point p=10.
  • For 0 < p < 10, R'(p) is positive, so R(p) is increasing.
  • For p > 10, R'(p) is negative, so R(p) is decreasing. This confirms that revenue is maximized at p=10, precisely the point of unit elasticity.

Differentiation provides the essential tools to analyze dynamic change and optimize outcomes, forming the analytical core of modern business and economic decision-making.

2.9 Chapter 2 Practice Problems

These problems allow you to practice the core skills of differentiation. Problems 1-2 cover the essential rules. Problems 3-11 apply the derivative to analyze rates of change, while problems 12-28 focus on approximation with differentials. Problems 29-30 introduce marginal analysis, 31-33 test the chain rule, 34-38 involve higher-order derivatives, and 39-43 explore the economic concept of elasticity.

1. Differentiate the following functions, simplifying the results. a. y = x² + 3x + 3 b. f(x) = x⁹ + 8x⁵ + 12x c. y = x² + 1/x – 1/x² d. f(x) = x³ + 1/x³ e. f(x) = (x+1)(3x-2) f. f(x) = (x²+5)(1-2x) g. f(x) = (100-x)(2x+1) h. y = (20-4x)(2x²+1) i. f(x) = (x³+1)(2x²-5x+1) j. f(x) = (x³-5x²+2)(x⁴-4) k. y = (x-5)/(x+4) l. f(x) = x³/(x³+3)
2. Find the rate of change of f(x) with respect to x for the prescribed value of x. a. f(x) = x³ – 3x + 5, x=2 b. f(x) = √x + 5x, x=4 c. f(x) = (x²+2)(x+√x), x=4 d. f(x) = (x²+3)(5-2x³), x=1 e. f(x) = (x-1)/(x+1), x=1 f. f(x) = x³/(2x-4), x=0
3. It is estimated that t years from now, the circulation of a local newspaper will be C(t) = 100t² + 400t + 5,000. a. Derive an expression for the rate at which the circulation will be changing with respect to time t years from now. b. At what rate will the circulation be changing with respect to time 5 years from now? c. By how much will the circulation actually change during the 6th year?
4. An efficiency study indicates that an average worker will have assembled f(x) = -x³ + 6x² + 15x radios x hours after starting work at 8:00 A.M. a. Derive a formula for the rate at which the worker will be assembling radios after x hours. b. At what rate will the worker be assembling radios at 9:00 A.M.? c. How many radios will the worker actually assemble between 9:00 and 10:00 A.M.?
5. Find the percentage rate of change in the function f(t) = 3t² – 7t + 5 with respect to t, when t=2.
6. Find the percentage rate of change in the function f(x) = x(x+3)² with respect to x, when x=3.
7. It is projected that x months from now, the population of a certain town will be P(x) = 2x³ + 4x² + 5,000. a. At what rate will the population be changing with respect to time 9 months from now? b. At what percentage rate will the population be changing with respect to time 9 months from now?
8. The gross annual earnings of a certain company were A(t) = 0.1t² + 10t + 20 thousand dollars t years after its formation in 1987. a. At what rate were the gross annual earnings of the company growing with respect to time in 1991? b. At what percentage rate were the gross annual earnings growing in 1991?
9. Records indicate that x years after 1985, the average property tax on a three-bedroom home in a certain community was T(x) = 20x² + 40x + 600. a. At what rate was the property tax increasing with respect to time in 1991? b. At what percentage rate was the property tax increasing in 1991?
10. It is estimated that t years from now, the population of a certain town will be P(t) = 20t + 10,000 / (t+2). a. Express the percentage rate of change of the population as a function of t. b. What will happen to the percentage rate of change of the population in the long run?
11. The gross national product (GNP) of a certain country is growing at a constant rate. In 1986 the GNP was 125 billion dollars, and in 1988 the GNP was 155 billion dollars. At what percentage rate was the GNP growing in 1991?
12. Estimate how much the function f(x) = x² – 3x + 5 will change as x increases from 5 to 5.3.
13. Estimate how much the function f(x) = x/(x+1) – 3 will change as x decreases from 4 to 3.8.
14. Estimate the percentage change in the function f(x) = x² + 2x – 9 as x increases from 4 to 4.3.
15. Estimate the percentage change in the function f(x) = 3x + 2/x as x decreases from 5 to 4.6.
16. A manufacturer’s total cost is C(q) = 0.1q³ – 0.5q² + 500q + 200 dollars when the level of production is q units. The current level of production is 4 units, and the manufacturer is planning to increase this to 4.1 units. Estimate how the total cost will change as a result.
17. It is projected that t years from now, the circulation of a local newspaper will be C(t) = 100t² + 400t + 5,000. Estimate the amount by which the circulation will increase during the next 6 months.
18. An environmental study suggests that t years from now, the average level of carbon monoxide in the air will be Q(t) = 0.05t² + 0.1t + 3.4 parts per million. By approximately how much will the carbon monoxide level change during the coming 6 months?
19. An efficiency study indicates that an average worker will have assembled f(x) = -x³ + 6x² + 15x radios x hours later. Approximately how many radios will the worker assemble between 9:00 and 9:15 A.M.?
20. At a certain factory, the daily output is Q(k) = 600k¹/² units, where k denotes the capital investment. The current capital investment is $900,000. Estimate the effect that an additional capital investment of $800 will have on the daily output.
21. At a certain factory, the daily output is Q(L) = 60,000L¹/³ units, where L denotes the size of the labor force. Currently 1000 worker-hours of labor are used each day. Estimate the effect on output that will be produced if the labor force is cut to 940 worker-hours.
22. At a certain factory, the daily output is Q(K,L) = 3000K¹/²L¹/³ where K denotes the firm’s capital investment and L denotes the size of the labor force. Suppose that the current capital investment is $400,000 and that 1,331 worker-hours of labor are used each day. Use marginal analysis to estimate the effect that an additional capital investment of $1,000 will have on the daily output if the size of the labor force is not changed.
23. The daily output at a certain factory is Q(L) = 300L²/³ units, where L denotes the size of the labor force. Currently 512 worker-hours of labor are used each day. Estimate the number of additional worker-hours of labor that will be needed to increase daily output by 12.5 units.
24. A manufacturer’s total cost is C(q) = (1/6)q³ + 642q + 400 dollars when q units are produced. The current level of production is 4 units. Estimate the amount by which the manufacturer should decrease production to reduce the total cost by $130.
25. Records indicate that x years after 1988, the average property tax on a three-bedroom home in a certain community was T(x) = 60x³/² + 40x + 1,200 dollars. Estimate the percentage by which the property tax increased during the first half of 1992.
26. The output at a certain factory is Q(K) = 400K¹/² units, where K denotes the firm’s capital investment. Estimate the percentage increase in output that will result from a 1 percent increase in capital investment.
27. The output at a certain factory is Q(K,L) = 600K¹/²L¹/³. Estimate the percentage increase in output that will result from a 2 percent increase in the size of the labor force if capital investment is not changed.
28. At a certain factory, the daily output is Q(K) = 1,200K¹/² units. Estimate the percentage increase in capital investment that is needed to produce a 1.2 percent increase in output.
29. Suppose the total cost of manufacturing q units is C(q) = 3q² + q + 500. a. Use marginal analysis to estimate the cost of manufacturing the 41st unit. b. Compute the actual cost of manufacturing the 41st unit.
30. A manufacturer’s total cost is C(q) = 0.1q³ – 0.5q² + 500q + 200. a. Use marginal analysis to estimate the cost of manufacturing the 4th unit. b. Compute the actual cost of manufacturing the 4th unit.
31. Use the chain rule to compute dy/dx. a. y=u²+1, u=3x-2 b. y=2u²-u+5, u=1-x² c. y=√u, u=x²-2x+3 d. y=1/u, u=x²+1 e. y=1/u², u=x²-9 f. y=u², u=1/(x-1)
32. Differentiate the following functions. a. f(x) = √(x⁴+1) b. f(x) = (1/(x-1))⁴ c. f(x) = √(x²+1)³
33. The gross annual earnings of a company were f(t) = √(10t²+t+236) thousand dollars t years after its formation in January 1988. a. At what rate were the gross annual earnings growing in January 1992? b. At what percentage rate were the gross annual earnings growing in January 1992?
34. An efficiency study indicates that an average worker will have produced Q(t) = -t³+8t²+15t units t hours later. a. Compute the worker’s rate of production at 9:00 A.M. b. At what rate is the worker’s rate of production changing at 9:00 A.M.? c. Use calculus to estimate the change in the worker’s rate of production between 9:00 and 9:15 A.M. d. Compute the actual change in the worker’s rate of production between 9:00 and 9:15 A.M.
35. Suppose the total cost of manufacturing q units of a commodity is C(q) = 3q²+5q+75. a. At what level of production is the average cost per unit the smallest? b. At what level of production is the average cost per unit equal to the marginal cost?
36. The problem is the same as that in problem 35 for C(q) = q³+5q+162.
37. Suppose the total revenue from the sale of q units is R(q) = -2q²+68q-128. a. At what level of sales is the average revenue per unit equal to the marginal revenue? b. Verify that average revenue is increasing if the level of sales is less than the level in part a. and decreasing if the level of sales is greater.
38. Assume that total national consumption is C(x) = 8 – 0.8x + 0.8√x. Find the marginal propensity to consume and determine the value of x that results in the smallest total savings.
39. Suppose the demand equation for a commodity is q = 60 – 0.1p. a. Express the elasticity of demand as a function of p. b. Calculate the elasticity when p=200. Interpret the answer. c. At what price is the elasticity of demand equal to -1?
40. Suppose the demand equation for a commodity is q = 200 – 2p². a. Express the elasticity of demand as a function of p. b. Calculate the elasticity when p=6. Interpret the answer. c. At what price is the elasticity of demand equal to -1?
41. Suppose the demand equation is q = 500 – 2p. a. Determine where the demand is elastic, inelastic, and of unit elasticity. b. Use the results from part a. to determine the intervals of increase and decrease of the revenue function. c. Find the total revenue function and use its first derivative to determine its intervals of increase and decrease and the price at which revenue is maximized.
42. Suppose the demand equation is q = 120 – 0.1p². a. Determine where the demand is elastic, inelastic, and of unit elasticity. b. Use the results from part a. to determine the intervals of increase and decrease of the revenue function. c. Find the total revenue function and use its first derivative to determine its intervals of increase and decrease and the price at which revenue is maximized.
43. Suppose demand q and price p are related by p = 60 – 2q. a. Express the elasticity of demand as a function of q. b. Calculate the elasticity when q=10. Interpret the answer. c. Substitute for q to express elasticity as a function of p. d. Use the original definition of η to express elasticity as a function of p.