AP Calculus AB Quiz: Derivative Rules Of Constant Sum Difference
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Derivative Rules Of Constant Sum DifferenceQuestion 1 of 20

If f(x)=5x3−2x2+x−9f(x) = 5x^3 - 2x^2 + x - 9, what is f′(x)f'(x)?

15x2−4x+115x^2 - 4x + 1
15x2−4x15x^2 - 4x
5x2−2x+15x^2 - 2x + 1
15x2−4x+1−915x^2 - 4x + 1 - 9
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AP Calculus AB Quiz

AP Calculus AB Quiz: Derivative Rules Of Constant Sum Difference

Practice Derivative Rules Of Constant Sum Difference in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Derivative Rules Of Constant Sum Difference, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

If f(x)=5x3−2x2+x−9f(x) = 5x^3 - 2x^2 + x - 9, what is f′(x)f'(x)?

  1. 15x2−4x+115x^2 - 4x + 1 (correct answer)
  2. 15x2−4x15x^2 - 4x
  3. 5x2−2x+15x^2 - 2x + 1
  4. 15x2−4x+1−915x^2 - 4x + 1 - 9

Explanation: To find the derivative of f(x)f(x), we apply the power rule, sum/difference rule, and constant rule to each term. The derivative of 5x35x^3 is 15x215x^2. The derivative of −2x2-2x^2 is −4x-4x. The derivative of xx is 11. The derivative of the constant −9-9 is 00. Combining these gives f′(x)=15x2−4x+1f'(x) = 15x^2 - 4x + 1.

Question 2

If h(x)=f(x)+g(x)h(x) = f(x) + g(x), where ff and gg are differentiable functions, and the line tangent to the graph of ff at x=3x=3 is y=2x−1y = 2x - 1 and the line tangent to the graph of gg at x=3x=3 is y=−x+7y = -x + 7, what is h′(3)h'(3)?

  1. 11 (correct answer)
  2. 33
  3. 66
  4. 99

Explanation: The derivative of a function at a point is the slope of the tangent line at that point. From the equation of the tangent line to ff, y=2x−1y = 2x - 1, the slope is 2. So, f′(3)=2f'(3) = 2. From the equation of the tangent line to gg, y=−x+7y = -x + 7, the slope is -1. So, g′(3)=−1g'(3) = -1. Using the sum rule for derivatives, h′(x)=f′(x)+g′(x)h'(x) = f'(x) + g'(x). Therefore, h′(3)=f′(3)+g′(3)=2+(−1)=1h'(3) = f'(3) + g'(3) = 2 + (-1) = 1.

Question 3

Let f(t)=9t3−2t2−6t+1f(t)=9t^3-2t^2-6t+1. What is f′(t)f'(t)?

  1. 27t2−4t−627t^2-4t-6 (correct answer)
  2. 27t3−4t2−6t27t^3-4t^2-6t
  3. 9t3−2t2−6t+19t^3-2t^2-6t+1
  4. 9t2−2t−69t^2-2t-6
  5. 27t2−4t−527t^2-4t-5

Explanation: Linearity rules allow us to handle the sum and differences in f(t) by differentiating individually with the constant multiple rule. For f(t) = 9t^3 - 2t^2 - 6t + 1, apply the power rule to every term. The derivative of 9t^3 is 27t^2, of -2t^2 is -4t, of -6t is -6, and of 1 is 0. Therefore, f'(t) = 27t^2 - 4t - 6, corresponding to choice A. A common rule misuse is applying the power rule without multiplying by the exponent, e.g., t^3 to t^2 instead of 3t^2. Another is ignoring negative signs in differences. A transferable strategy is to process each term with d/dx (c xnx^n) = c n x^{n-1}, remember constants vanish, and maintain sign integrity in the final expression.

Question 4

Let f(x)=3x2−7f(x) = 3x^2 - 7 and g(x)=5x+2g(x) = 5x + 2. If h(x)=2f(x)−3g(x)+4h(x) = 2f(x) - 3g(x) + 4, what is h′(x)h'(x)?

  1. 12x−1512x - 15 (correct answer)
  2. 6x−156x - 15
  3. 12x−1112x - 11
  4. 6x−116x - 11

Explanation: Using the constant multiple, sum, and difference rules: h(x)=2(3x2−7)−3(5x+2)+4=6x2−14−15x−6+4=6x2−15x−16h(x) = 2(3x^2 - 7) - 3(5x + 2) + 4 = 6x^2 - 14 - 15x - 6 + 4 = 6x^2 - 15x - 16. Therefore, h′(x)=12x−15h'(x) = 12x - 15. Choice B incorrectly uses coefficient 6 instead of 12 for the xx term. Choice C forgets the −3g(x)-3g(x) term in the constant. Choice D makes both errors.

Question 5

If F(x)=αf(x)+βg(x)+γF(x) = \alpha f(x) + \beta g(x) + \gamma where α\alpha, β\beta, γ\gamma are nonzero constants, and it's known that F′(x)=0F'(x) = 0 for all xx, which of the following must be true?

  1. f(x)=g(x)=0f(x) = g(x) = 0 for all xx
  2. f′(x)=g′(x)=0f'(x) = g'(x) = 0 for all xx
  3. αf′(x)+βg′(x)=0\alpha f'(x) + \beta g'(x) = 0 for all xx (correct answer)
  4. α=β=γ=0\alpha = \beta = \gamma = 0

Explanation: Since F′(x)=αf′(x)+βg′(x)+0=αf′(x)+βg′(x)F'(x) = \alpha f'(x) + \beta g'(x) + 0 = \alpha f'(x) + \beta g'(x) (the derivative of constant γ\gamma is 0), and F′(x)=0F'(x) = 0 for all xx, we must have αf′(x)+βg′(x)=0\alpha f'(x) + \beta g'(x) = 0 for all xx. Choice A confuses functions with their derivatives. Choice B is too restrictive; the individual derivatives need not be zero. Choice D contradicts the given that the constants are nonzero.

Question 6

If y=2x2−πx+e+13y = \sqrt{2}x^2 - \pi x + e + \frac{1}{\sqrt{3}}, then dydx\frac{dy}{dx} equals:

  1. 22x−π+1232\sqrt{2}x - \pi + \frac{1}{2\sqrt{3}}
  2. 22x−π+e2\sqrt{2}x - \pi + e
  3. 22x−π2\sqrt{2}x - \pi (correct answer)
  4. 2x−π+e\sqrt{2}x - \pi + e

Explanation: Using the power rule and constant rule: ddx[2x2]=2⋅2x=22x\frac{d}{dx}[\sqrt{2}x^2] = \sqrt{2} \cdot 2x = 2\sqrt{2}x, ddx[−πx]=−π\frac{d}{dx}[-\pi x] = -\pi, and ddx[e+13]=0\frac{d}{dx}[e + \frac{1}{\sqrt{3}}] = 0 since both ee and 13\frac{1}{\sqrt{3}} are constants. Choice A incorrectly differentiates the constant 13\frac{1}{\sqrt{3}}. Choice B incorrectly keeps ee as non-zero. Choice D uses wrong coefficient for x2x^2 term and keeps constant ee.

Question 7

Let F(x)=3x3−4x2+5x−6F(x)=3x^3-4x^2+5x-6. What is the derivative F′(x)F'(x)?

  1. 9x2−8x+59x^2-8x+5 (correct answer)
  2. 9x3−8x2+5x9x^3-8x^2+5x
  3. 3x3−4x2+5x−63x^3-4x^2+5x-6
  4. 3x2−8x+53x^2-8x+5
  5. 9x2−8x−69x^2-8x-6

Explanation: Linearity rules allow independent differentiation of terms in F(x) = 3x^3 - 4x^2 + 5x - 6. Derivative of 3x^3 is 9x^2, of -4x^2 is -8x, of 5x is 5, of -6 is 0. Thus, F'(x) = 9x^2 - 8x + 5. A common misuse is subtracting instead of adding derivatives in sums. Constants are often incorrectly included. Decompose using linearity and apply power rule to each term for accuracy.

Question 8

Let G(x)=−6x2+13x−15G(x)= -6x^2+13x-15. What is the derivative G′(x)G'(x)?

  1. −12x+13-12x+13 (correct answer)
  2. −12x2+13x-12x^2+13x
  3. −6x2+13x−15-6x^2+13x-15
  4. −6x+13-6x+13
  5. −12x−15-12x-15

Explanation: The sum and difference rules of linearity allow term-by-term differentiation of polynomials. For G(x) = -6x^2 + 13x - 15, derivative of -6x^2 is -12x, of 13x is 13, and of -15 is 0. Thus, G'(x) = -12x + 13. A common misuse is including the constant's derivative as nonzero, like adding -15. Signs on linear terms must be preserved accurately. For quadratic functions, use linearity to differentiate each power separately, ensuring constants are ignored.

Question 9

A function is y(t)=13t2+4t+9y(t)=13t^2+4t+9. What is y′(t)y'(t)?

  1. 26t+426t+4 (correct answer)
  2. 26t2+4t26t^2+4t
  3. 13t2+4t+913t^2+4t+9
  4. 13t+413t+4
  5. 26t+926t+9

Explanation: Linearity rules facilitate term-by-term differentiation for y(t) = 13t^2 + 4t + 9. Derivative of 13t^2 is 26t, of 4t is 4, of +9 is 0. Thus, y'(t) = 26t + 4. A common misuse is including the constant in the derivative, like adding 9. Linear terms are sometimes overlooked. For quadratics, apply sum rules and power rule individually to each term.

Question 10

Let N(x)=5x4+7x3−14x+9N(x)=5x^4+7x^3-14x+9. What is N′(x)N'(x)?

  1. 20x3+21x2−1420x^3+21x^2-14 (correct answer)
  2. 20x4+21x3−14x20x^4+21x^3-14x
  3. 5x4+7x3−14x+95x^4+7x^3-14x+9
  4. 5x3+21x2−145x^3+21x^2-14
  5. 20x3+21x2+920x^3+21x^2+9

Explanation: Linearity rules facilitate term-by-term differentiation for N(x) with sums and multiples. For N(x) = 5x^4 + 7x^3 - 14x + 9, use the power rule. The derivative of 5x^4 is 20x^3, of 7x^3 is 21x^2, of -14x is -14, and of 9 is 0. Therefore, N'(x) = 20x^3 + 21x^2 - 14, which is choice A. A common misuse is forgetting multiplication for cubic terms. Some include constants. A transferable strategy is to sequence terms, apply d/dx (c xnx^n) = c n x^{n-1}, and combine using linearity.

Question 11

The position of a particle moving along the x-axis is given by p(t)=t3−6t2+5p(t) = t^3 - 6t^2 + 5 for time t≥0t \ge 0. What is the velocity of the particle at the instant its acceleration is zero?

  1. −12-12 (correct answer)
  2. 22
  3. −11-11
  4. 00

Explanation: The velocity function is the first derivative of the position function: v(t)=p′(t)=3t2−12tv(t) = p'(t) = 3t^2 - 12t. The acceleration function is the second derivative of the position function: a(t)=p′′(t)=6t−12a(t) = p''(t) = 6t - 12. To find when the acceleration is zero, set a(t)=0a(t) = 0: 6t−12=06t - 12 = 0, which gives t=2t = 2. To find the velocity at this time, substitute t=2t=2 into the velocity function: v(2)=3(2)2−12(2)=3(4)−24=12−24=−12v(2) = 3(2)^2 - 12(2) = 3(4) - 24 = 12 - 24 = -12.

Question 12

What is the derivative of f(x)=(2x−3)2f(x) = (2x - 3)^2?

  1. 8x−128x - 12 (correct answer)
  2. 2(2x−3)2(2x - 3)
  3. 4x−64x - 6
  4. 8x8x

Explanation: First, expand the expression for f(x)f(x): f(x)=(2x−3)(2x−3)=4x2−6x−6x+9=4x2−12x+9f(x) = (2x - 3)(2x - 3) = 4x^2 - 6x - 6x + 9 = 4x^2 - 12x + 9. Now, differentiate the expanded polynomial term by term: f'(x) = rac{d}{dx}(4x^2 - 12x + 9) = 8x - 12.

Question 13

A rate function is r(t)=−9t4+10t2+7t−3r(t)= -9t^4+10t^2+7t-3. What is r′(t)r'(t)?

  1. −36t3+20t+7-36t^3+20t+7 (correct answer)
  2. −36t4+20t2+7t-36t^4+20t^2+7t
  3. −9t4+10t2+7t−3-9t^4+10t^2+7t-3
  4. −9t3+20t+7-9t^3+20t+7
  5. −36t3+20t−3-36t^3+20t-3

Explanation: The rate function uses linearity rules to differentiate terms independently with differences and multiples. For r(t) = -9t^4 + 10t^2 + 7t - 3, use the power rule. The derivative of -9t^4 is -36t^3, of 10t^2 is 20t, of 7t is 7, and of -3 is 0. Therefore, r'(t) = -36t^3 + 20t + 7, which is choice A. Common misuse is forgetting to multiply by the power for even degrees. Some keep the constant. A transferable strategy is to apply d/dx (c xnx^n) = c n x^{n-1} to each, preserve signs from differences, and ignore constants in the sum.

Question 14

For S(x)=10x5−5x4+4x3−1S(x)=10x^5-5x^4+4x^3-1, what is S′(x)S'(x)?

  1. 50x4−20x3+12x250x^4-20x^3+12x^2 (correct answer)
  2. 50x5−20x4+12x350x^5-20x^4+12x^3
  3. 10x5−5x4+4x3−110x^5-5x^4+4x^3-1
  4. 10x4−20x3+12x210x^4-20x^3+12x^2
  5. 50x4−5x3+12x250x^4-5x^3+12x^2

Explanation: The linearity of differentiation allows us to compute the derivative of a polynomial by differentiating each term separately and respecting constant multiples. For S(x) = 10x^5 - 5x^4 + 4x^3 - 1, apply the power rule to each term: the derivative of ax^n is a n x^{n-1}. Thus, the derivative of 10x^5 is 50x^4, of -5x^4 is -20x^3, of 4x^3 is 12x^2, and the constant -1 has derivative 0. Therefore, S'(x) = 50x^4 - 20x^3 + 12x^2. A common misuse is forgetting that the derivative of a constant is zero, leading to incorrectly including terms like -1 in the derivative. Another error is mishandling the constant multiple, such as not multiplying the coefficient by the exponent. For differentiating polynomials, always break down into individual terms, apply the power rule with linearity, and confirm constants vanish.

Question 15

Let y(x)=11x9+3x5−x+8y(x)=11x^9+3x^5-x+8. What is y′(x)y'(x)?

  1. 99x8+15x4−199x^8+15x^4-1 (correct answer)
  2. 99x9+15x5−x99x^9+15x^5-x
  3. 11x9+3x5−x+811x^9+3x^5-x+8
  4. 11x8+15x4−111x^8+15x^4-1
  5. 99x8+15x5−199x^8+15x^5-1

Explanation: Linearity of differentiation supports separate term handling for y(x) with sum and constant multiple rules. For y(x) = 11x^9 + 3x^5 - x + 8, apply the power rule. The derivative of 11x^9 is 99x^8, of 3x^5 is 15x^4, of -x is -1, and of 8 is 0. So, y'(x) = 99x^8 + 15x^4 - 1, matching choice A. A common rule misuse is incorrect coefficient calculation for high powers like 11*9=99. Another is deriving -x as 0. A transferable strategy is to compute large coefficients accurately, apply d/dx (c xnx^n) = c n x^{n-1}, and assemble using linearity.

Question 16

Let f(x)=−2x4−9x3+6x2−1f(x)= -2x^4-9x^3+6x^2-1. What is f′(x)f'(x)?

  1. −8x3−27x2+12x-8x^3-27x^2+12x (correct answer)
  2. −8x4−27x3+12x2-8x^4-27x^3+12x^2
  3. −2x4−9x3+6x2−1-2x^4-9x^3+6x^2-1
  4. −2x3−27x2+12x-2x^3-27x^2+12x
  5. −8x3−27x2−1-8x^3-27x^2-1

Explanation: Using linearity, compute f'(x) for f(x) = -2x^4 - 9x^3 + 6x^2 - 1 by differentiating each term. Derivative of -2x^4 is -8x^3, of -9x^3 is -27x^2, of 6x^2 is 12x, of -1 is 0. Thus, f'(x) = -8x^3 - 27x^2 + 12x. A common misuse is ignoring negative signs, turning terms positive incorrectly. Constants are often mistakenly differentiated. Break polynomials into sums and use power rule per term with linearity for reliable outcomes.

Question 17

For m(t)=6t2−15t+9m(t)=6t^2-15t+9, what is the derivative m′(t)m'(t)?

  1. 12t−1512t-15 (correct answer)
  2. 12t2−15t12t^2-15t
  3. 6t2−15t+96t^2-15t+9
  4. 6t−156t-15
  5. 12t−612t-6

Explanation: Linearity rules allow independent differentiation of terms in m(t) using sum, difference, and constant multiple. For m(t) = 6t^2 - 15t + 9, apply the power rule. The derivative of 6t^2 is 12t, of -15t is -15, and of 9 is 0. Therefore, m'(t) = 12t - 15, which is choice A. A common misuse is treating quadratic terms as linear without multiplying by 2. Another error is deriving the constant as non-zero. A transferable strategy is to focus on low-degree polynomials by applying d/dx (c xnx^n) = c n x^{n-1} term-wise and combining via linearity.

Question 18

A profit function is P(x)=−x5+13x3−4x+6P(x)=-x^5+13x^3-4x+6. What is P′(x)P'(x)?

  1. −5x4+39x2−4-5x^4+39x^2-4 (correct answer)
  2. −5x5+39x3−4x-5x^5+39x^3-4x
  3. −x5+13x3−4x+6-x^5+13x^3-4x+6
  4. −x4+13x2−4-x^4+13x^2-4
  5. −5x4+39x2+6-5x^4+39x^2+6

Explanation: The profit function's derivative employs linearity to handle sums and differences term by term with constant multiples. For P(x) = -x^5 + 13x^3 - 4x + 6, use the power rule. The derivative of -x^5 is -5x^4, of 13x^3 is 39x^2, of -4x is -4, and of 6 is 0. So, P'(x) = -5x^4 + 39x^2 - 4, matching choice A. Common misuse is ignoring negative signs, turning -x^5 to 5x^4. Some forget to derive linear terms correctly. A transferable rule-selection strategy is to note all coefficients, apply d/dx (c xnx^n) = c n x^{n-1}, ensure signs are preserved, and eliminate constants.

Question 19

A volume model is V(x)=15x2−6x+20V(x)=15x^2-6x+20. What is V′(x)V'(x)?

  1. 30x−630x-6 (correct answer)
  2. 30x2−6x30x^2-6x
  3. 15x2−6x+2015x^2-6x+20
  4. 15x−615x-6
  5. 30x+2030x+20

Explanation: The volume model's derivative uses linearity to treat terms separately with differences and multiples. For V(x) = 15x^2 - 6x + 20, apply the power rule. The derivative of 15x^2 is 30x, of -6x is -6, and of 20 is 0. Thus, V'(x) = 30x - 6, corresponding to choice A. Common misuse includes not multiplying by 2 for quadratic terms. Another is deriving constants. A transferable strategy is to focus on quadratic forms, use d/dx (c xnx^n) = c n x^{n-1}, and assemble with original operations.

Question 20

A function is B(t)=−11t3+5t−8B(t)= -11t^3+5t-8. What is B′(t)B'(t)?

  1. −33t2+5-33t^2+5 (correct answer)
  2. −33t3+5t-33t^3+5t
  3. −11t3+5t−8-11t^3+5t-8
  4. −11t2+5-11t^2+5
  5. −33t2−8-33t^2-8

Explanation: The constant multiple and sum rules under linearity simplify differentiating B(t)=−11t3+5t−8B(t) = -11t^3 + 5t - 8. Derivative of −11t3-11t^3 is −33t2-33t^2, of 5t5t is 55, of −8-8 is 00. Thus, B′(t)=−33t2+5B'(t) = -33t^2 + 5. A common misuse is treating the constant as having a derivative, like −8-8 becoming −8t-8t or similar. Negative coefficients need careful handling to avoid sign flips. For such functions, use linearity to isolate terms and apply power rule accurately.