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)=x4−8x3+10x−1f(x) = x^4 - 8x^3 + 10x - 1. What is f′′(x)f''(x)?

  1. 12x2−48x12x^2 - 48x (correct answer)
  2. 4x3−24x2+104x^3 - 24x^2 + 10
  3. 24x−4824x - 48
  4. 12x2−48x+1012x^2 - 48x + 10
Explanation: To find the second derivative, we first find the first derivative. f′(x)=4x3−24x2+10f'(x) = 4x^3 - 24x^2 + 10. Then, we differentiate f′(x)f'(x) to find f′′(x)f''(x). f′′(x)=ddx(4x3−24x2+10)=12x2−48xf''(x) = \frac{d}{dx}(4x^3 - 24x^2 + 10) = 12x^2 - 48x.

Question 5

If f(x)=4sin⁡(x)−ex+1f(x) = 4\sin(x) - e^x + 1, what is the value of f′(0)f'(0)?

  1. 33 (correct answer)
  2. 44
  3. −1-1
  4. 00
Explanation: First, find the derivative of the function: f′(x)=ddx(4sin⁡(x)−ex+1)=4cos⁡(x)−exf'(x) = \frac{d}{dx}(4\sin(x) - e^x + 1) = 4\cos(x) - e^x. Next, evaluate the derivative at x=0x=0: f′(0)=4cos⁡(0)−e0=4(1)−1=3f'(0) = 4\cos(0) - e^0 = 4(1) - 1 = 3.

Question 6

What is the slope of the line tangent to the graph of y=4x5−7x3+2y = 4x^5 - 7x^3 + 2 at x=−1x = -1?

  1. −1-1 (correct answer)
  2. 4141
  3. 11
  4. −41-41
Explanation: The slope of the tangent line is the value of the derivative at the given point. First, find the derivative: dydx=ddx(4x5−7x3+2)=20x4−21x2\frac{dy}{dx} = \frac{d}{dx}(4x^5 - 7x^3 + 2) = 20x^4 - 21x^2. Now, evaluate the derivative at x=−1x = -1: 20(−1)4−21(−1)2=20(1)−21(1)=20−21=−120(-1)^4 - 21(-1)^2 = 20(1) - 21(1) = 20 - 21 = -1.

Question 7

If g(x)=4x3−8x2xg(x) = \frac{4x^3 - 8x}{2x} for x≠0x \neq 0, what is g′(x)g'(x)?

  1. 4x4x (correct answer)
  2. 2x−42x - 4
  3. 4x−44x - 4
  4. 12x2−82\frac{12x^2 - 8}{2}
Explanation: For x≠0x \neq 0, we can simplify the expression for g(x)g(x) before differentiating. g(x)=4x32x−8x2x=2x2−4g(x) = \frac{4x^3}{2x} - \frac{8x}{2x} = 2x^2 - 4. Now, we can easily find the derivative: g′(x)=ddx(2x2−4)=4xg'(x) = \frac{d}{dx}(2x^2 - 4) = 4x.

Question 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

The function f(x)=k1g1(x)+k2g2(x)+k3g3(x)f(x) = k_1g_1(x) + k_2g_2(x) + k_3g_3(x) where k1=3k_1 = 3, k2=−2k_2 = -2, k3=5k_3 = 5, g1′(x)=4xg_1'(x) = 4x, g2′(x)=−x2g_2'(x) = -x^2, and g3′(x)=7g_3'(x) = 7. What is f′(x)f'(x)?

  1. 12x+2x2−3512x + 2x^2 - 35
  2. 12x+2x2+3512x + 2x^2 + 35 (correct answer)
  3. 12x−2x2+3512x - 2x^2 + 35
  4. 12x−2x2−3512x - 2x^2 - 35
Explanation: When you encounter a function that's a linear combination of other functions, you're dealing with one of calculus's most fundamental properties: the linearity of derivatives. The derivative of a sum equals the sum of derivatives, and constants can be factored out. To find f′(x)f'(x), you need to differentiate each term separately. Since f(x)=3g1(x)−2g2(x)+5g3(x)f(x) = 3g_1(x) - 2g_2(x) + 5g_3(x), we have: f′(x)=3g1′(x)−2g2′(x)+5g3′(x)f'(x) = 3g_1'(x) - 2g_2'(x) + 5g_3'(x) Substituting the given derivatives: f′(x)=3(4x)−2(−x2)+5(7)f'(x) = 3(4x) - 2(-x^2) + 5(7) f′(x)=12x+2x2+35f'(x) = 12x + 2x^2 + 35 This matches answer choice B. Let's examine why the other options are incorrect: Choice A (12x+2x2−3512x + 2x^2 - 35) incorrectly makes the constant term negative, likely from mishandling k3⋅g3′(x)=5⋅7=35k_3 \cdot g_3'(x) = 5 \cdot 7 = 35. Choice C (12x−2x2+3512x - 2x^2 + 35) makes the x2x^2 term negative, suggesting confusion about the double negative: k2⋅g2′(x)=(−2)⋅(−x2)=+2x2k_2 \cdot g_2'(x) = (-2) \cdot (-x^2) = +2x^2. Choice D (12x−2x2−3512x - 2x^2 - 35) combines both errors from A and C, getting both the x2x^2 and constant terms wrong. Study tip: When differentiating linear combinations, work systematically through each term and pay extra attention to signs. Double negatives are common sources of error—when you see (−2)×(−x2)(-2) \times (-x^2), remember that negative times negative equals positive.

Question 16

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 17

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)=ddx(4x2−12x+9)=8x−12f'(x) = \frac{d}{dx}(4x^2 - 12x + 9) = 8x - 12.

Question 18

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 19

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 20

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.