Day 21 - Quotient Rule Notes

Course Information and Learning Objectives

  • Course Module: Day 21 - Quotient Rule

  • Textbook Reference: Section 11.311.3

  • Core Learning Objectives:

    • Apply the quotient rule to compute derivatives of rational functions and complex algebraic quotients.

    • Formulate and interpret the average cost function Cˉ(x)\bar{C}(x) in economic contexts.

    • Analyze the relationship between marginal cost C′(x)C'(x) and average cost Cˉ(x)\bar{C}(x) to establish conditions for increasing, decreasing, and minimal unit cost.

The Quotient Rule for Differentiation

  • Formal Definition: For differentiable functions f(x)f(x) and g(x)g(x), the derivative of their quotient f(x)g(x)\frac{f(x)}{g(x)} is defined wherever the denominator g(x)≠0g(x) \neq 0:

    • ddx[f(x)g(x)]=f′(x)g(x)−f(x)g′(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

  • Operational Guidelines:

    • Numerator: (Derivative of top) ×\times (Original bottom) −- (Original top) ×\times (Derivative of bottom).

    • Denominator: Square of the original bottom function: [g(x)]2[g(x)]^2.

    • Order of Subtraction: Order is rigid; reversing terms introduces an erroneous sign change.

Worked Examples: Applying the Quotient Rule
Example 21.1(a): Rational Polynomial Function

  • Function: y=3x3+x−5x+1y = \frac{3x^3 + x - 5}{x + 1}

  • Component Derivatives:

    • Numerator: f(x)=3x3+x−5  ⟹  f′(x)=9x2+1f(x) = 3x^3 + x - 5 \implies f'(x) = 9x^2 + 1

    • Denominator: g(x)=x+1  ⟹  g′(x)=1g(x) = x + 1 \implies g'(x) = 1

  • Quotient Rule Application:

    • y′=(9x2+1)(x+1)−(3x3+x−5)(1)(x+1)2y' = \frac{(9x^2 + 1)(x + 1) - (3x^3 + x - 5)(1)}{(x + 1)^2}

  • Expansion & Simplification:

    • Numerator expansion: (9x3+9x2+x+1)−(3x3+x−5)(9x^3 + 9x^2 + x + 1) - (3x^3 + x - 5)

    • Combine like terms: 6x3+9x2+66x^3 + 9x^2 + 6

  • Final Answer:

    • Standard form: y′=6x3+9x2+6(x+1)2y' = \frac{6x^3 + 9x^2 + 6}{(x + 1)^2}

    • Factored form: y′=3(2x3+3x2+2)(x+1)2y' = \frac{3(2x^3 + 3x^2 + 2)}{(x + 1)^2}

Example 21.1(b): Quotient with Factored Denominator

  • Function: g(t)=10t(t2+1)(3t−1)g(t) = \frac{10t}{(t^2 + 1)(3t - 1)}

  • Component Derivatives:

    • Numerator: u(t)=10t  ⟹  u′(t)=10u(t) = 10t \implies u'(t) = 10

    • Denominator expanded: v(t)=3t3−t2+3t−1  ⟹  v′(t)=9t2−2t+3v(t) = 3t^3 - t^2 + 3t - 1 \implies v'(t) = 9t^2 - 2t + 3

  • Quotient Rule Application:

    • g′(t)=10(3t3−t2+3t−1)−10t(9t2−2t+3)[(t2+1)(3t−1)]2g'(t) = \frac{10(3t^3 - t^2 + 3t - 1) - 10t(9t^2 - 2t + 3)}{[(t^2 + 1)(3t - 1)]^2}

  • Numerator Simplification:

    • Expand: (30t3−10t2+30t−10)−(90t3−20t2+30t)=−60t3+10t2−10(30t^3 - 10t^2 + 30t - 10) - (90t^3 - 20t^2 + 30t) = -60t^3 + 10t^2 - 10

  • Final Answer:

    • Standard form: g′(t)=−60t3+10t2−10(t2+1)2(3t−1)2g'(t) = \frac{-60t^3 + 10t^2 - 10}{(t^2 + 1)^2(3t - 1)^2}

    • Factored form: g′(t)=−10(6t3−t2+1)(t2+1)2(3t−1)2g'(t) = \frac{-10(6t^3 - t^2 + 1)}{(t^2 + 1)^2(3t - 1)^2}

The Average Cost Function and Marginal Analysis

  • Definitions:

    • Total Cost: C(x)C(x) is the cost of producing xx units.

    • Average Cost: Cˉ(x)=C(x)x\bar{C}(x) = \frac{C(x)}{x}, representing unit cost.

  • Derivation of Marginal Average Cost:

    • Cˉ′(x)=ddx[C(x)x]=xC′(x)−C(x)(1)x2=C′(x)−C(x)xx=C′(x)−Cˉ(x)x\bar{C}'(x) = \frac{d}{dx}\left[\frac{C(x)}{x}\right] = \frac{x C'(x) - C(x)(1)}{x^2} = \frac{C'(x) - \frac{C(x)}{x}}{x} = \frac{C'(x) - \bar{C}(x)}{x}

  • Economic Criteria (for physical output x>0x > 0):

    • Increasing Average Cost: Cˉ′(x)>0  ⟺  C′(x)>Cˉ(x)\bar{C}'(x) > 0 \iff C'(x) > \bar{C}(x) (Marginal cost exceeds average cost).

    • Decreasing Average Cost: Cˉ′(x)<0  ⟺  C′(x)<Cˉ(x)\bar{C}'(x) < 0 \iff C'(x) < \bar{C}(x) (Marginal cost is below average cost).

    • Optimum / Minimum Average Cost: Cˉ′(x)=0  ⟺  C′(x)=Cˉ(x)\bar{C}'(x) = 0 \iff C'(x) = \bar{C}(x) (Average cost is minimized where marginal cost equals average cost).

Real-World Applications: Marginal vs. Average Cost in Production
Example 21.2: Automotive Production Rate Analysis

  • Scenario: Manufacturing firm scaling output at x=3x = 3 thousand cars (3 0003\,000 cars).

    • Marginal Cost: C′(3)=12 000 dollars/carC'(3) = 12\,000\text{ dollars/car}

    • Average Cost: Cˉ(3)=40 000 dollars/car\bar{C}(3) = 40\,000\text{ dollars/car}

  • Computation:

    • Per thousand cars (x=3x = 3):

    • Cˉ′(3)=12 000−40 0003=−28 0003≈−9 333.33 dollars per car per thousand cars\bar{C}'(3) = \frac{12\,000 - 40\,000}{3} = \frac{-28\,000}{3} \approx -9\,333.33\text{ dollars per car per thousand cars}

    • Per individual car (x=3 000x = 3\,000):

    • Cˉ′(3 000)=12 000−40 0003 000=−283≈−9.33 dollars per car per car\bar{C}'(3\,000) = \frac{12\,000 - 40\,000}{3\,000} = -\frac{28}{3} \approx -9.33\text{ dollars per car per car}

  • Interpretation: Because marginal cost (12 00012\,000) is significantly less than average cost (40 00040\,000), producing additional vehicles drives unit cost downward.

Logistic Growth Modeling: Video Game Sales
Example 21.3: Cyberpunk II Sales Velocity

  • Model: Sales q(t)=10 0001+0.5(32)−tq(t) = \frac{10\,000}{1 + 0.5\left(\frac{3}{2}\right)^{-t}} units at tt months.

  • Target Time: Two years   ⟹  t=24\implies t = 24 months.

  • Differentiation:

    • Numerator: u(t)=10 000  ⟹  u′(t)=0u(t) = 10\,000 \implies u'(t) = 0

    • Denominator: v(t)=1+0.5(32)−t  ⟹  v′(t)=−0.5ln⁡(32)(32)−tv(t) = 1 + 0.5\left(\frac{3}{2}\right)^{-t} \implies v'(t) = -0.5\ln\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)^{-t}

    • q′(t)=−10 000[−0.5ln⁡(32)(32)−t][1+0.5(32)−t]2=5 000ln⁡(32)(32)−t[1+0.5(32)−t]2q'(t) = \frac{-10\,000\left[-0.5\ln\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)^{-t}\right]}{\left[1 + 0.5\left(\frac{3}{2}\right)^{-t}\right]^2} = \frac{5\,000\ln\left(\frac{3}{2}\right)\left(\frac{3}{2}\right)^{-t}}{\left[1 + 0.5\left(\frac{3}{2}\right)^{-t}\right]^2}

  • Evaluation at t=24t = 24:

    • ln⁡(1.5)≈0.405465\ln(1.5) \approx 0.405465

    • (1.5)−24≈5.945×10−5(1.5)^{-24} \approx 5.945 \times 10^{-5}

    • Numerator: 5 000×0.405465×5.945×10−5≈0.120535\,000 \times 0.405465 \times 5.945 \times 10^{-5} \approx 0.12053

    • Denominator: [1+0.5(5.945×10−5)]2≈1.00006\left[1 + 0.5(5.945 \times 10^{-5})\right]^2 \approx 1.00006

    • q′(24)≈0.12 units per monthq'(24) \approx 0.12\text{ units per month}

  • Conclusion: Sales have reached carrying capacity, stabilizing at approximately 0.12 units/month0.12\text{ units/month}.

Practice Exercises and Step-by-Step Solutions
You Try Exercise (a): Polynomial Quotient

  • Function: y=x2+3x−5x3−1y = \frac{x^2 + 3x - 5}{x^3 - 1}

  • Derivatives: f′(x)=2x+3f'(x) = 2x + 3, g′(x)=3x2g'(x) = 3x^2

  • Expansion:

    • y′=(2x+3)(x3−1)−(x2+3x−5)(3x2)(x3−1)2y' = \frac{(2x + 3)(x^3 - 1) - (x^2 + 3x - 5)(3x^2)}{(x^3 - 1)^2}

    • Numerator: (2x4+3x3−2x−3)−(3x4+9x3−15x2)(2x^4 + 3x^3 - 2x - 3) - (3x^4 + 9x^3 - 15x^2)

  • Result:

    • y′=−x4−6x3+15x2−2x−3(x3−1)2y' = \frac{-x^4 - 6x^3 + 15x^2 - 2x - 3}{(x^3 - 1)^2}

You Try Exercise (b): Constant Numerator with Cubic Denominator

  • Function: g(t)=5(5t+1)(3t2−t+1)g(t) = \frac{5}{(5t + 1)(3t^2 - t + 1)}

  • Expanded Denominator: D(t)=15t3−2t2+4t+1D(t) = 15t^3 - 2t^2 + 4t + 1

  • Derivatives: N′(t)=0N'(t) = 0, D′(t)=45t2−4t+4D'(t) = 45t^2 - 4t + 4

  • Quotient Rule:

    • g′(t)=0−5(45t2−4t+4)[(5t+1)(3t2−t+1)]2g'(t) = \frac{0 - 5(45t^2 - 4t + 4)}{[(5t + 1)(3t^2 - t + 1)]^2}

  • Result:

    • g′(t)=−225t2+20t−20(5t+1)2(3t2−t+1)2g'(t) = \frac{-225t^2 + 20t - 20}{(5t + 1)^2(3t^2 - t + 1)^2}

Key Terminology and Deliverables

  • Quotient Rule: [fg]′=f′g−fg′g2\left[\frac{f}{g}\right]' = \frac{f'g - fg'}{g^2}

  • Average Cost: Cˉ(x)=C(x)x\bar{C}(x) = \frac{C(x)}{x}

  • Marginal Cost: C′(x)=ddx[C(x)]C'(x) = \frac{d}{dx}[C(x)]

  • Course Deliverables: HW 9 due next Thursday