Calculus Study Notes: Product Rule, Quotient Rule, and Chain Rule

Section 1.6: The Product Rule

  • Fundamental Rule Principle: The derivative of a product of 2 functions is NOT the product of their derivatives.

  • Formal Definition of the Product Rule:

    • Let F(x)=f(x)⋅g(x)F(x) = f(x) \cdot g(x).

    • The derivative F′(x)F'(x) is defined as:     F′(x)=ddx[f(x)⋅g(x)]=f′(x)⋅g(x)+g′(x)⋅f(x)F'(x) = \frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + g'(x) \cdot f(x)

    • Alternative notations:

    • y=f⋅g  ⟹  y′=f′g+g′fy = f \cdot g \implies y' = f'g + g'f

    • F′(x)=g′(x)⋅f(x)+f′(x)⋅g(x)F'(x) = g'(x) \cdot f(x) + f'(x) \cdot g(x)

  • Worked Examples:

    • Example 1: Find the derivative of y=(2x5+x−1)(3x−2)y = (2x^5 + x - 1)(3x - 2)

    • Identify component functions and their derivatives:

      • f(x)=2x5+x−1  ⟹  f′(x)=10x4+1f(x) = 2x^5 + x - 1 \implies f'(x) = 10x^4 + 1

      • g(x)=3x−2  ⟹  g′(x)=3g(x) = 3x - 2 \implies g'(x) = 3

    • Apply the Product Rule formula y′=f′g+g′fy' = f'g + g'f:       y′=(10x4+1)(3x−2)+3(2x5+x−1)y' = (10x^4 + 1)(3x - 2) + 3(2x^5 + x - 1)

    • Example 2: Find the derivative of y=(x+1)(x−x)y = (\sqrt{x} + 1)(\sqrt{x} - x)

    • Express radicals as fractional exponents:

      • f(x)=x+1=x1/2+1  ⟹  f′(x)=12x−1/2f(x) = \sqrt{x} + 1 = x^{1/2} + 1 \implies f'(x) = \frac{1}{2}x^{-1/2}

      • g(x)=x−x=x1/2−x  ⟹  g′(x)=12x−1/2−1g(x) = \sqrt{x} - x = x^{1/2} - x \implies g'(x) = \frac{1}{2}x^{-1/2} - 1

    • Apply the Product Rule formula y′=f′g+g′fy' = f'g + g'f:       y′=(12x−1/2)(x−x)+(12x−1/2−1)(x+1)y' = \left(\frac{1}{2}x^{-1/2}\right)(\sqrt{x} - x) + \left(\frac{1}{2}x^{-1/2} - 1\right)(\sqrt{x} + 1)

    • Example 3: Find the derivative of y=5x−3(x4−5x3+10x−2)y = 5x^{-3}(x^4 - 5x^3 + 10x - 2)

    • Identify component functions and their derivatives:

      • f(x)=5x−3  ⟹  f′(x)=5(−3)x−3−1=−15x−4f(x) = 5x^{-3} \implies f'(x) = 5(-3)x^{-3-1} = -15x^{-4}

      • g(x)=x4−5x3+10x−2  ⟹  g′(x)=4x3−15x2+10g(x) = x^4 - 5x^3 + 10x - 2 \implies g'(x) = 4x^3 - 15x^2 + 10

    • Apply the Product Rule formula y′=f′g+g′fy' = f'g + g'f:       y′=(−15x−4)(x4−5x3+10x−2)+(4x3−15x2+10)(5x−3)y' = (-15x^{-4})(x^4 - 5x^3 + 10x - 2) + (4x^3 - 15x^2 + 10)(5x^{-3})

    • Note: No simplification is required for this result.

Section 1.6: The Quotient Rule

  • Fundamental Rule Principle: The derivative of a quotient (fraction) function is NOT the quotient of the derivatives.

  • Formal Definition of the Quotient Rule:

    • Let Q(x)=f(x)g(x)Q(x) = \frac{f(x)}{g(x)}.

    • The derivative Q′(x)Q'(x) is defined as:     Q′(x)=ddx[f(x)g(x)]=g(x)⋅f′(x)−f(x)⋅g′(x)[g(x)]2Q'(x) = \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}

    • Alternative notation:     y′=f′g−g′fg2y' = \frac{f'g - g'f}{g^2}

    • Mnemonic Device for Remembering the Quotient Rule:

    • Let f(x)=Top=Hif(x) = \text{Top} = \text{Hi} and g(x)=Bottom=Log(x) = \text{Bottom} = \text{Lo}

    • Formula memory aid:       Lo⋅d(Hi)−Hi⋅d(Lo)LoLo\frac{\text{Lo} \cdot d(\text{Hi}) - \text{Hi} \cdot d(\text{Lo})}{\text{LoLo}}

  • Worked Examples:

    • Example 1: Find the derivative of y=1−3xx2+2y = \frac{1 - 3x}{x^2 + 2}

    • Identify numerator and denominator derivatives:

      • f(x)=1−3x  ⟹  f′(x)=−3f(x) = 1 - 3x \implies f'(x) = -3

      • g(x)=x2+2  ⟹  g′(x)=2xg(x) = x^2 + 2 \implies g'(x) = 2x

    • Apply Quotient Rule y′=f′g−g′fg2y' = \frac{f'g - g'f}{g^2}:       y′=(−3)(x2+2)−(2x)(1−3x)(x2+2)2y' = \frac{(-3)(x^2 + 2) - (2x)(1 - 3x)}{(x^2 + 2)^2}

    • Simplify the numerator:       −3(x2+2)−2x(1−3x)=−3x2−6−2x+6x2=3x2−2x−6-3(x^2 + 2) - 2x(1 - 3x) = -3x^2 - 6 - 2x + 6x^2 = 3x^2 - 2x - 6

    • Final simplified derivative:       y′=3x2−2x−6(x2+2)2y' = \frac{3x^2 - 2x - 6}{(x^2 + 2)^2}

    • Example 2: Find the derivative of y=3x4+2xx3−1y = \frac{3x^4 + 2x}{x^3 - 1}

    • Identify component derivatives:

      • f′(x)=12x3+2f'(x) = 12x^3 + 2

      • g′(x)=3x2g'(x) = 3x^2

    • Apply Quotient Rule y′=f′g−g′fg2y' = \frac{f'g - g'f}{g^2}:       y′=(12x3+2)(x3−1)−(3x2)(3x4+2x)(x3−1)2y' = \frac{(12x^3 + 2)(x^3 - 1) - (3x^2)(3x^4 + 2x)}{(x^3 - 1)^2}

    • Expand numerator terms:       (12x3+2)(x3−1)=12x6−12x3+2x3−2=12x6−10x3−2(12x^3 + 2)(x^3 - 1) = 12x^6 - 12x^3 + 2x^3 - 2 = 12x^6 - 10x^3 - 2       (3x2)(3x4+2x)=9x6+6x3(3x^2)(3x^4 + 2x) = 9x^6 + 6x^3

    • Combine numerator terms:       12x6−12x3+2x3−2−9x6−6x3=3x6−16x3−212x^6 - 12x^3 + 2x^3 - 2 - 9x^6 - 6x^3 = 3x^6 - 16x^3 - 2

    • Final simplified derivative:       y′=3x6−16x3−2(x3−1)2y' = \frac{3x^6 - 16x^3 - 2}{(x^3 - 1)^2}

Application Example: Narcotic Concentration in Bloodstream

  • Problem Statement: The amount of a narcotic in milligrams remaining in a patient's bloodstream tt hours after administration can be modeled by:   N(t)=150.2t+1N(t) = \frac{15}{0.2t + 1}, where 0≤t≤150 \le t \le 15

  • Evaluation and Interpretation Problems:

    • Part a: Evaluate N(4)N(4) and interpret.

    • Calculation:       N(4)=150.2(4)+1=150.8+1=151.8=8.33 mgN(4) = \frac{15}{0.2(4) + 1} = \frac{15}{0.8 + 1} = \frac{15}{1.8} = 8.33\,mg

    • Interpretation: 4 hrs4\,hrs after administration, 8.33 mg8.33\,mg of narcotic remains in the patient's bloodstream.

    • Part b: Evaluate N′(4)N'(4) and interpret.

    • Find derivative N′(t)N'(t) using Quotient Rule:

      • f(t)=15  ⟹  f′(t)=0f(t) = 15 \implies f'(t) = 0

      • g(t)=0.2t+1  ⟹  g′(t)=0.2g(t) = 0.2t + 1 \implies g'(t) = 0.2

      • N′(t)=f′g−g′fg2=(0)(0.2t+1)−(0.2)(15)(0.2t+1)2=−3(0.2t+1)2N'(t) = \frac{f'g - g'f}{g^2} = \frac{(0)(0.2t + 1) - (0.2)(15)}{(0.2t + 1)^2} = \frac{-3}{(0.2t + 1)^2}

    • Evaluate at t=4t = 4:       N′(4)=−3(0.2(4)+1)2=−3(1.8)2=−0.93 mg/hrN'(4) = \frac{-3}{(0.2(4) + 1)^2} = \frac{-3}{(1.8)^2} = -0.93\,mg/hr

    • Interpretation: The amount of narcotic remaining in the patient's bloodstream is decreasing by 0.93 mg/hr0.93\,mg/hr at 4 hrs4\,hrs after administration.

    • Part c: At what rate is the amount of the narcotic decreasing initially?

    • Initial rate corresponds to t=0t = 0:       N′(0)=−3(0.2(0)+1)2=−3(1)2=−3 mg/hrN'(0) = \frac{-3}{(0.2(0) + 1)^2} = \frac{-3}{(1)^2} = -3\,mg/hr

    • Interpretation: Initially (at time t=0t = 0), the amount of narcotic is decreasing at a rate of 3 mg/hr3\,mg/hr

    • Part d: When will the amount of the narcotic be decreasing by 0.48 mg/hr0.48\,mg/hr?

    • Definition: Find tt such that N′(t)=−0.48 mg/hrN'(t) = -0.48\,mg/hr.

    • Calculator Intersection Method:

      • Input Y1=N′(t)=−3(0.2t+1)2Y_1 = N'(t) = \frac{-3}{(0.2t + 1)^2}

      • Input Y2=−0.48Y_2 = -0.48

      • Window Settings: Xmin=0X_{min} = 0, Xmax=15X_{max} = 15 (given domain 0≤t≤150 \le t \le 15)

      • Find the intersection point of Y1Y_1 and Y2Y_2

    • Result: At t=1.5 hrst = 1.5\,hrs after administration, the amount of narcotic is decreasing by 0.48 mg/hr0.48\,mg/hr

Section 1.7: The Chain Rule and Extended Power Rule

  • Motivational Introductory Example:

    • Consider the function f(x)=(1+x2)2f(x) = (1 + x^2)^2. Find the derivative.

  • Definition of The Extended Power Rule:

    • Let g(x)g(x) be any differentiable function of xx. Then for any real number kk:     ddx[[g(x)]k]=k⋅[g(x)]k−1⋅g′(x)\frac{d}{dx}\left[[g(x)]^k\right] = k \cdot [g(x)]^{k-1} \cdot g'(x)

  • Revisiting Introductory Example with Extended Power Rule:

    • Function: f(x)=(1+x2)2f(x) = (1 + x^2)^2

    • Components: g(x)=1+x2g(x) = 1 + x^2, k=2k = 2, g′(x)=2xg'(x) = 2x

    • Calculation steps:     f′(x)=2(1+x2)2−1⋅(2x)=4(1+x2)1⋅x=4x(1+x2)=4x+4x3f'(x) = 2(1 + x^2)^{2-1} \cdot (2x) = 4(1 + x^2)^1 \cdot x = 4x(1 + x^2) = 4x + 4x^3


Extended Power Rule example

  • Worked Examples:

    • Example 1: Find the derivative of y=(x4+2x2+1)3y = (x^4 + 2x^2 + 1)^3

    • Identify inner function and derivative:

      • g(x)=x4+2x2+1g(x) = x^4 + 2x^2 + 1

      • g′(x)=4x3+4xg'(x) = 4x^3 + 4x

    • Apply Extended Power Rule:       y′=3(x4+2x2+1)2⋅(4x3+4x)y' = 3(x^4 + 2x^2 + 1)^2 \cdot (4x^3 + 4x)

    • Explicit Rule: There is no need to factor this expression further!

    • Example 2: Find the derivative of y=1+8xy = \sqrt{1 + 8x}

    • Rewrite radical as fractional power:       y=(1+8x)1/2y = (1 + 8x)^{1/2}

    • Identify inner function derivative:       g′(x)=8g'(x) = 8

    • Apply Extended Power Rule:       y′=12(1+8x)1/2−1⋅(8)=4(1+8x)−1/2=4(1+8x)1/2=41+8xy' = \frac{1}{2}(1 + 8x)^{1/2 - 1} \cdot (8) = 4(1 + 8x)^{-1/2} = \frac{4}{(1 + 8x)^{1/2}} = \frac{4}{\sqrt{1 + 8x}}

    • Example 3: Find the derivative of y=2x2−1(3x4+2)2y = \frac{2x^2 - 1}{(3x^4 + 2)^2}

    • Combination of Quotient Rule and Extended Power Rule:

      • Numerator: f(x)=2x2−1  ⟹  f′(x)=4xf(x) = 2x^2 - 1 \implies f'(x) = 4x

      • Denominator: g(x)=(3x4+2)2  ⟹  g′(x)=2(3x4+2)1⋅(12x3)=24x3(3x4+2)g(x) = (3x^4 + 2)^2 \implies g'(x) = 2(3x^4 + 2)^1 \cdot (12x^3) = 24x^3(3x^4 + 2)

    • Apply Quotient Rule y′=f′g−g′fg2y' = \frac{f'g - g'f}{g^2}:       y′=(4x)(3x4+2)2−24x3(3x4+2)(2x2−1)[(3x4+2)2]2y' = \frac{(4x)(3x^4 + 2)^2 - 24x^3(3x^4 + 2)(2x^2 - 1)}{[(3x^4 + 2)^2]^2}

    • Step-by-step simplification:

      • Factor out common term 4x(3x4+2)4x(3x^4 + 2) from numerator:         y′=4x(3x4+2)[(3x4+2)−6x2(2x2−1)](3x4+2)4y' = \frac{4x(3x^4 + 2)[(3x^4 + 2) - 6x^2(2x^2 - 1)]}{(3x^4 + 2)^4}

      • Expand inner bracket:         (3x4+2)−6x2(2x2−1)=3x4+2−12x4+6x2=−9x4+6x2+2(3x^4 + 2) - 6x^2(2x^2 - 1) = 3x^4 + 2 - 12x^4 + 6x^2 = -9x^4 + 6x^2 + 2

      • Cancel common factor (3x4+2)(3x^4 + 2) in numerator and denominator:         y′=4x(−9x4+6x2+2)(3x4+2)3y' = \frac{4x(-9x^4 + 6x^2 + 2)}{(3x^4 + 2)^3}

      • Distribute 4x4x across numerator terms:         y′=−36x5+24x3+8x(3x4+2)3y' = \frac{-36x^5 + 24x^3 + 8x}{(3x^4 + 2)^3}

    • Example 4: Find the derivative of y=(x+5)7(4x−1)10y = (x + 5)^7(4x - 1)^{10}

    • Combination of Product Rule and Extended Power Rule:

      • First term derivative: f=(x+5)7  ⟹  f′=7(x+5)7−1⋅(1)=7(x+5)6f = (x + 5)^7 \implies f' = 7(x + 5)^{7-1} \cdot (1) = 7(x + 5)^6

      • Second term derivative: g=(4x−1)10  ⟹  g′=10(4x−1)10−1⋅(4)=40(4x−1)9g = (4x - 1)^{10} \implies g' = 10(4x - 1)^{10-1} \cdot (4) = 40(4x - 1)^9

    • Apply Product Rule y′=f′g+g′fy' = f'g + g'f:       y′=7(x+5)6(4x−1)10+40(4x−1)9(x+5)7y' = 7(x + 5)^6(4x - 1)^{10} + 40(4x - 1)^9(x + 5)^7

    • Step-by-step factoring and simplification:

      • Factor out common binomial factors (x+5)6(4x−1)9(x + 5)^6(4x - 1)^9:         y′=(x+5)6(4x−1)9[7(4x−1)+40(x+5)]y' = (x + 5)^6(4x - 1)^9 [7(4x - 1) + 40(x + 5)]

      • Expand and collect terms within square brackets:         7(4x−1)+40(x+5)=28x−7+40x+200=68x+1937(4x - 1) + 40(x + 5) = 28x - 7 + 40x + 200 = 68x + 193

      • Final simplified derivative answer:         y′=(x+5)6(4x−1)9(68x+193)y' = (x + 5)^6(4x - 1)^9(68x + 193)