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Introduction to Derivatives · Ages 17–18

Differentiating with Product, Quotient, and Chain Rules

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Designed for 12th grade calculus students. Identify the primary rule required for each function, then compute the derivative.

  1. Identify the primary rule and find f'(x) for f(x) = (x^3 + 4)^5.

  2. Identify the primary rule and find f'(x) for f(x) = (x^2 + 1)(3x - 2).

  3. Identify the primary rule and find f'(x) for f(x) = (2x) / (x + 5).

  4. Identify the primary rule and find f'(x) for f(x) = 4 * sqrt(3x^2 - 1).

  5. Identify the rules needed and find f'(x) for f(x) = x^2 * (5x - 1)^4.

  6. Identify the rules needed and find f'(x) for f(x) = (3x + 2)^2 / (x - 1).

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Differentiating with Product, Quotient, and Chain Rules

Answer key

  1. Identify the primary rule and find f'(x) for f(x) = (x^3 + 4)^5.

    Chain rule. Outer derivative is 5(x^3 + 4)^4; inner derivative is 3x^2. Result: f'(x) = 15x^2(x^3 + 4)^4.

  2. Identify the primary rule and find f'(x) for f(x) = (x^2 + 1)(3x - 2).

    Product rule. Derivative is 2x(3x - 2) + (x^2 + 1)(3) = 6x^2 - 4x + 3x^2 + 3. Result: f'(x) = 9x^2 - 4x + 3.

  3. Identify the primary rule and find f'(x) for f(x) = (2x) / (x + 5).

    Quotient rule. Derivative is [2(x + 5) - 2x(1)] / (x + 5)^2 = [2x + 10 - 2x] / (x + 5)^2. Result: f'(x) = 10 / (x + 5)^2.

  4. Identify the primary rule and find f'(x) for f(x) = 4 * sqrt(3x^2 - 1).

    Chain rule. Rewrite as 4(3x^2 - 1)^(1/2). Outer derivative is 2(3x^2 - 1)^(-1/2); inner derivative is 6x. Result: f'(x) = 12x / sqrt(3x^2 - 1).

  5. Identify the rules needed and find f'(x) for f(x) = x^2 * (5x - 1)^4.

    Product rule with chain rule nested. Derivative is 2x(5x - 1)^4 + x^2 * [4(5x - 1)^3 * 5]. Factor out common terms. Result: f'(x) = 2x(5x - 1)^3(15x - 1).

  6. Identify the rules needed and find f'(x) for f(x) = (3x + 2)^2 / (x - 1).

    Quotient rule with chain rule in the numerator. Numerator derivative is 2(3x + 2)(3) = 6(3x + 2). Result: f'(x) = [6(3x + 2)(x - 1) - (3x + 2)^2] / (x - 1)^2.