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Math Grade 9-12 Answer Key

Calculus: Derivative Rules: Power, Chain, Product, Quotient

Practice finding derivatives with the main differentiation rules

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Calculus: Derivative Rules: Power, Chain, Product, Quotient

Practice finding derivatives with the main differentiation rules

Math - Grade 9-12

Instructions: Read each problem carefully. Show your work, identify the rule you use when helpful, and simplify your answer when possible.
  1. 1

    Find d/dx of x^7.

    Use d/dx of x^n = nx^(n - 1).

    The derivative is 7x^6 by the power rule.
  2. 2

    Let f(x) = 5x^4 - 3x^2 + 7. Find f'(x).

    The derivative is f'(x) = 20x^3 - 6x. The constant 7 has derivative 0.
  3. 3

    Find the derivative of y = 4x^(-3) + sqrt(x).

    Rewrite sqrt(x) as x^(1/2) before differentiating.

    The derivative is y' = -12x^(-4) + 1/(2sqrt(x)). This uses the power rule with sqrt(x) written as x^(1/2).
  4. 4

    Find the derivative of y = (3x^2 - 1)^5.

    Use the chain rule: derivative of the outside times derivative of the inside.

    The derivative is y' = 30x(3x^2 - 1)^4. The outside derivative gives 5(3x^2 - 1)^4, and the inside derivative is 6x.
  5. 5

    Find the derivative of y = sqrt(2x + 5).

    The derivative is y' = 1/sqrt(2x + 5). This comes from writing the function as (2x + 5)^(1/2) and applying the chain rule.
  6. 6

    Let y = (x^2 + 1)(x^3 - 4x). Find y' using the product rule.

    For y = uv, use y' = u'v + uv'.

    The derivative is y' = 2x(x^3 - 4x) + (x^2 + 1)(3x^2 - 4), which simplifies to y' = 5x^4 - 9x^2 - 4.
  7. 7

    Find the derivative of y = x^2 sin(x).

    The derivative is y' = 2x sin(x) + x^2 cos(x). This uses the product rule with u = x^2 and v = sin(x).
  8. 8

    Let y = (2x + 1)/(x - 3). Find y' using the quotient rule.

    For y = u/v, use y' = (u'v - uv')/v^2.

    The derivative is y' = -7/(x - 3)^2. The numerator becomes 2(x - 3) - (2x + 1), which simplifies to -7.
  9. 9

    Find the derivative of y = (x^2 + 4)/(3x^2 - 1).

    The derivative is y' = -26x/(3x^2 - 1)^2. This comes from applying the quotient rule and simplifying the numerator.
  10. 10

    Let y = (x^3 - 2x)^4. Find y' and then evaluate y' at x = 1.

    First use the chain rule, then substitute x = 1 into the derivative.

    The derivative is y' = 4(x^3 - 2x)^3(3x^2 - 2). At x = 1, y' = -4.
  11. 11

    Let y = (x^2 + 1)(3x - 2)^4. Find y'.

    The derivative is y' = 2x(3x - 2)^4 + 12(x^2 + 1)(3x - 2)^3. This uses the product rule and the chain rule.
  12. 12

    Find the derivative of y = (x^2 - 1)/(x^2 + 1).

    Keep the denominator squared until after simplifying the numerator.

    The derivative is y' = 4x/(x^2 + 1)^2. The quotient rule gives a numerator of 2x(x^2 + 1) - 2x(x^2 - 1), which simplifies to 4x.
  13. 13

    Find the equation of the tangent line to f(x) = x^3 - 4x at x = 2.

    Use the derivative for the slope and the original function for the point on the graph.

    The derivative is f'(x) = 3x^2 - 4, so the slope at x = 2 is 8. Since f(2) = 0, the tangent line is y = 8x - 16.
  14. 14

    A student says the derivative of y = (x^2 + 1)^3 is y' = 3(x^2 + 1)^2. Identify the mistake and give the correct derivative.

    The student forgot to multiply by the derivative of the inside function, x^2 + 1. The correct derivative is y' = 6x(x^2 + 1)^2.
  15. 15

    The position of an object is s(t) = t^4 - 2t^3 + t. Find the velocity v(t), the acceleration a(t), and their values at t = 2.

    Velocity is the first derivative of position, and acceleration is the derivative of velocity.

    The velocity is v(t) = 4t^3 - 6t^2 + 1, and the acceleration is a(t) = 12t^2 - 12t. At t = 2, the velocity is 9 and the acceleration is 24.
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