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

A company's cost is C(q)=12q6−8q+3C(q)=\frac{1}{2}q^6-8q+3 dollars; what is C′(q)C'(q)?

3q5−83q^5-8
12q6−8\frac{1}{2}q^6-8
3q5−8+33q^5-8+3
6q5−86q^5-8
3q5+83q^5+8
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AP Calculus BC Quiz

AP Calculus BC Quiz: Derivative Rules Of Constant Sum Difference

Practice Derivative Rules Of Constant Sum Difference in AP Calculus BC 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 BC.

How to use this quiz

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.

All questions

Question 1

A company's cost is C(q)=12q6−8q+3C(q)=\frac{1}{2}q^6-8q+3 dollars; what is C′(q)C'(q)?

  1. 3q5−83q^5-8 (correct answer)
  2. 12q6−8\frac{1}{2}q^6-8
  3. 3q5−8+33q^5-8+3
  4. 6q5−86q^5-8
  5. 3q5+83q^5+8
Explanation: This problem tests the basic derivative rules for constants, sums, and differences. To find C'(q) for C(q) = (1/2)q^6 - 8q + 3, apply the power rule to the first term: the derivative of (1/2)q^6 is (1/2)6q^{5} = 3q^5. For the second term, the derivative of -8q is -81q^{0} = -8. The constant term +3 has a derivative of 0, so it vanishes. A tempting distractor is choice C, which adds back +3, but constants are eliminated in differentiation. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

Question 2

The position of a cart is s(x)=−5x3+4x−9s(x)=-5x^3+4x-9 meters; find the velocity s′(x)s'(x).

  1. −15x2+4-15x^2+4 (correct answer)
  2. −15x2+4−9-15x^2+4-9
  3. −5x3+4-5x^3+4
  4. 15x2+415x^2+4
  5. −15x2−4-15x^2-4
Explanation: This problem tests the basic derivative rules for constants, sums, and differences. To find s'(x) for s(x)=−5x3+4x−9s(x) = -5x^3 + 4x - 9, apply the power rule to the first term: the derivative of −5x3-5x^3 is −5∗3∗x2=−15x2-5*3*x^{2} = -15x^2. For the second term, the derivative of +4x+4x is 4∗1∗x0=44*1*x^{0} = 4. The constant term −9-9 has a derivative of 0, so it is omitted from the result. A tempting distractor is choice B, which subtracts the constant −9-9 again, but constants do not persist in derivatives. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

Question 3

A particle's height is h(t)=9−2t5+6t2h(t)=9-2t^5+6t^2 meters; determine h′(t)h'(t).

  1. −10t4+12t-10t^4+12t (correct answer)
  2. 10t4+12t10t^4+12t
  3. −10t4+12t+9-10t^4+12t+9
  4. −2t5+6t2-2t^5+6t^2
  5. −10t4+6t-10t^4+6t
Explanation: This problem tests the basic derivative rules for constants, sums, and differences. To find h'(t) for h(t) = 9 - 2t^5 + 6t^2, apply the power rule to the second term: the derivative of -2t^5 is -25t^{4} = -10t^4. For the third term, the derivative of +6t^2 is 62t^{1} = 12t. The constant term 9 has a derivative of 0, so it is not included. A tempting distractor is choice C, which adds +9, but constants disappear upon differentiation. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

Question 4

Let f(x) = cos(x) and let g be a twice-differentiable function with g''(π\pi) = 4. If k(x) = 3g(x) - f(x) + 5x^2, what is the value of k''(π\pi)?

  1. 23
  2. 21 (correct answer)
  3. 11
  4. 10
Explanation: When you encounter a problem asking for the second derivative of a combination of functions, you need to apply differentiation rules systematically to each term. To find k′′(π)k''(\pi) where k(x)=3g(x)−f(x)+5x2k(x) = 3g(x) - f(x) + 5x^2, start by taking the first derivative: k′(x)=3g′(x)−f′(x)+10xk'(x) = 3g'(x) - f'(x) + 10x. Then take the second derivative: k′′(x)=3g′′(x)−f′′(x)+10k''(x) = 3g''(x) - f''(x) + 10. Now substitute x=πx = \pi: k′′(π)=3g′′(π)−f′′(π)+10k''(\pi) = 3g''(\pi) - f''(\pi) + 10. You're given that g′′(π)=4g''(\pi) = 4. Since f(x)=cos⁡(x)f(x) = \cos(x), we have f′(x)=−sin⁡(x)f'(x) = -\sin(x) and f′′(x)=−cos⁡(x)f''(x) = -\cos(x), so f′′(π)=−cos⁡(π)=−(−1)=1f''(\pi) = -\cos(\pi) = -(-1) = 1. Therefore: k′′(π)=3(4)−1+10=12−1+10=21k''(\pi) = 3(4) - 1 + 10 = 12 - 1 + 10 = 21. Choice A (23) likely results from incorrectly calculating f′′(π)f''(\pi) as 3 instead of 1, perhaps confusing the derivatives of sine and cosine. Choice C (11) probably comes from forgetting the coefficient 3 in front of g′′(π)g''(\pi), giving 4−1+10−2=114 - 1 + 10 - 2 = 11. Choice D (10) might result from omitting the g′′(π)g''(\pi) term entirely and only calculating −f′′(π)+10=−1+10=11-f''(\pi) + 10 = -1 + 10 = 11, or making an arithmetic error. Study tip: Always write out each step of differentiation clearly, especially for composite expressions. Double-check your trigonometric derivatives—cosine becomes negative sine, then negative cosine.

Question 5

Revenue is modeled by R(x)=3x6−11x3+4xR(x)=3x^6-11x^3+4x. What is R′(x)R'(x)?

  1. 18x5−33x2+418x^5-33x^2+4 (correct answer)
  2. 18x5−33x2+4x18x^5-33x^2+4x
  3. 18x6−33x2+418x^6-33x^2+4
  4. 18x5+33x2+418x^5+33x^2+4
  5. 3x6−11x3+43x^6-11x^3+4
Explanation: This problem requires finding the marginal revenue using basic derivative rules. We differentiate each term separately: the derivative of 3x⁶ is 18x⁵ (3·6·x⁵), the derivative of -11x³ is -33x² (-11·3·x²), and the derivative of 4x is 4. Combining these gives R'(x) = 18x⁵ - 33x² + 4. Choice B incorrectly writes the last term as 4x instead of 4, failing to recognize that d/dx(4x) = 4, not 4x. The systematic approach is to apply the power rule to each term, remembering that linear terms become constants.

Question 6

A function is defined by f(x)=−9x4+7x3−5f(x)= -9x^4+7x^3-5. What is f′(x)f'(x)?

  1. −36x3+21x2-36x^3+21x^2 (correct answer)
  2. −36x3+21x2−5-36x^3+21x^2-5
  3. −36x4+21x2-36x^4+21x^2
  4. 36x3+21x236x^3+21x^2
  5. −9x4+7x3-9x^4+7x^3
Explanation: This problem requires finding f'(x) using basic derivative rules on a polynomial function. We differentiate each term: the derivative of -9x⁴ is -36x³ (multiply -9 by 4 and reduce the exponent), the derivative of 7x³ is 21x² (7·3·x²), and the derivative of the constant -5 is 0. Thus, f'(x) = -36x³ + 21x² + 0 = -36x³ + 21x². Choice B incorrectly includes the constant -5 in the derivative, but constants always have zero derivative. The key is to apply the power rule term by term, remembering that constant terms disappear.

Question 7

In a lab, output is Q(t)=−(t4)+8t3−16Q(t)=-(t^4)+8t^3-16; find the rate of change Q′(t)Q'(t).

  1. −4t3+24t2-4t^3+24t^2 (correct answer)
  2. 4t3+24t24t^3+24t^2
  3. −4t3−24t2-4t^3-24t^2
  4. −t4+8t3-t^4+8t^3
  5. −3t3+24t2-3t^3+24t^2
Explanation: This problem assesses your understanding of the basic derivative rules for constants, sums, and differences in polynomials. To compute Q'(t), differentiate each term separately using the power rule. The derivative of -t^4 is -1 times 4t^{3}, which is -4t^3. The derivative of +8t^3 is 8 times 3t^{2}, which is +24t^2, and the derivative of -16 is 0, so they combine to -4t^3 + 24t^2. A tempting distractor like choice D is -t^4 + 8t^3, but this fails because it doesn't apply the power rule at all. Always remember to apply the power rule to each term independently and set the derivative of constants to zero.

Question 8

A profit function is R(x)=−2x4+13x−20R(x)=-2x^4+13x-20 (dollars). What is R′(x)R'(x)?

  1. −8x3+13-8x^3+13 (correct answer)
  2. −8x3+13x-8x^3+13x
  3. 8x3+138x^3+13
  4. −2x4+13-2x^4+13
  5. −6x3+13-6x^3+13
Explanation: This problem involves finding the marginal profit by differentiating the profit function. Starting with R(x) = -2x⁴ + 13x - 20, we apply the power rule to each term. The derivative of -2x⁴ is -2·4x³ = -8x³, the derivative of 13x is 13·1 = 13, and the derivative of the constant -20 is 0. Thus, R'(x) = -8x³ + 13 + 0 = -8x³ + 13. A mistake would be to write 13x instead of 13 (choice B), forgetting that the derivative of x¹ is 1, not x. Remember that when differentiating x to the first power, the result is simply the coefficient.

Question 9

A beam's deflection is y(x)=35x5−10x2+8y(x)=\frac{3}{5}x^5-10x^2+8; determine y′(x)y'(x).

  1. 3x4−20x3x^4-20x (correct answer)
  2. 35x4−20x\frac{3}{5}x^4-20x
  3. 3x4+20x3x^4+20x
  4. 3x4−20x+83x^4-20x+8
  5. 155x4−10x\frac{15}{5}x^4-10x
Explanation: This problem tests the basic derivative rules for constants, sums, and differences. To find y'(x) for y(x) = (3/5)x^5 - 10x^2 + 8, apply the power rule to the first term: the derivative of (3/5)x^5 is (3/5)5x^{4} = 3x^4. For the second term, the derivative of -10x^2 is -102x^{1} = -20x. The constant term +8 has a derivative of 0, so it is not present. A tempting distractor is choice D, which includes +8, but constants are differentiated to zero. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

Question 10

A temperature model is T(t)=6t2−(t6)+15tT(t)=6t^2-(t^6)+15t. Find T′(t)T'(t).

  1. 12t−6t5+1512t-6t^5+15 (correct answer)
  2. 12t−6t512t-6t^5
  3. 12t+6t5+1512t+6t^5+15
  4. 6t−6t5+156t-6t^5+15
  5. 12t−t6+1512t-t^6+15
Explanation: This temperature model requires differentiating terms including one in parentheses. Given T(t) = 6t² - (t⁶) + 15t, we differentiate each term: 6t² becomes 6·2t = 12t; -(t⁶) becomes -6t⁵; and 15t becomes 15·1 = 15. Thus, T'(t) = 12t - 6t⁵ + 15. Choice B omits the constant term 15, incorrectly thinking the derivative of 15t is 0, but the derivative of 15t is 15. The parentheses around t⁶ don't change the differentiation—we still apply the power rule normally to get -6t⁵.

Question 11

A tank's water volume is modeled by V(t)=7t4−3t2+12V(t)=7t^4-3t^2+12. What is V′(t)V'(t)?

  1. 28t3−6t+1228t^3-6t+12
  2. 28t3−6t28t^3-6t (correct answer)
  3. 7t3−3t+127t^3-3t+12
  4. 28t3+6t28t^3+6t
  5. 28t4−6t228t^4-6t^2
Explanation: This problem requires applying basic derivative rules to find the rate of change of water volume. To find V'(t) from V(t) = 7t⁴ - 3t² + 12, we differentiate each term separately using the power rule. For 7t⁴, we get 7·4t³ = 28t³; for -3t², we get -3·2t = -6t; and the constant 12 becomes 0. Therefore, V'(t) = 28t³ - 6t + 0 = 28t³ - 6t. A common error would be keeping the constant term, giving 28t³ - 6t + 12 (choice A), but remember that the derivative of any constant is zero. The key strategy is to apply the power rule term by term: bring down the exponent as a coefficient and reduce the exponent by one.

Question 12

Let f and g be differentiable functions. The line y = 5x - 3 is tangent to the graph of f at x = 2. The function h(x) = 4f(x) - 2g(x) has a horizontal tangent at x = 2. What is the value of g'(2)?

  1. 5
  2. 7
  3. 10 (correct answer)
  4. 14
Explanation: When you encounter problems involving composite functions and their derivatives, think about how the derivative rules interact with given geometric information about tangent lines. Since the line y=5x−3y = 5x - 3 is tangent to ff at x=2x = 2, you know that f′(2)=5f'(2) = 5 (the slope of the tangent line). You can also find that f(2)=7f(2) = 7 by substituting x=2x = 2 into the tangent line equation. For h(x)=4f(x)−2g(x)h(x) = 4f(x) - 2g(x), use the linearity of derivatives: h′(x)=4f′(x)−2g′(x)h'(x) = 4f'(x) - 2g'(x). Since hh has a horizontal tangent at x=2x = 2, you know h′(2)=0h'(2) = 0. Setting up the equation: h′(2)=4f′(2)−2g′(2)=0h'(2) = 4f'(2) - 2g'(2) = 0. Substituting f′(2)=5f'(2) = 5: 4(5)−2g′(2)=04(5) - 2g'(2) = 0, which gives 20−2g′(2)=020 - 2g'(2) = 0. Solving: g′(2)=10g'(2) = 10. Looking at the wrong answers: Choice (A) gives 5, which incorrectly assumes g′(2)=f′(2)g'(2) = f'(2). Choice (B) gives 7, which might come from confusing g′(2)g'(2) with f(2)f(2). Choice (D) gives 14, which could result from sign errors or incorrect algebraic manipulation. The correct answer is (C) 10. Study tip: When working with composite functions and tangent line information, always write out the derivative formula first, then substitute known values systematically. Watch for horizontal tangents—they're a key signal that the derivative equals zero at that point.

Question 13

A runner's distance is d(x)=12x3+7x2−4x+1d(x)=12x^3+7x^2-4x+1; compute d′(x)d'(x).

  1. 36x2+14x−436x^2+14x-4 (correct answer)
  2. 36x2+14x−4+136x^2+14x-4+1
  3. 12x3+7x2−412x^3+7x^2-4
  4. 36x2−14x−436x^2-14x-4
  5. 24x2+14x−424x^2+14x-4
Explanation: This problem tests the basic derivative rules for constants, sums, and differences. To find d'(x) for d(x) = 12x^3 + 7x^2 - 4x + 1, apply the power rule to the first term: the derivative of 12x^3 is 123x^{2} = 36x^2. For the second term, the derivative of +7x^2 is 72x^{1} = 14x, and for -4x it is -41x^{0} = -4. The constant term +1 has a derivative of 0, so it drops out. A tempting distractor is choice B, which adds +1, but constants vanish in derivatives. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

Question 14

A tank's net water volume is modeled by V(t)=7t4−3t2+12V(t)=7t^4-3t^2+12 liters; what is V′(t)V'(t)?

  1. 28t3−6t+1228t^3-6t+12
  2. 28t3−6t28t^3-6t (correct answer)
  3. 7t4−3t27t^4-3t^2
  4. 28t3+6t28t^3+6t
  5. 14t3−6t14t^3-6t
Explanation: This problem tests the basic derivative rules for constants, sums, and differences. To find V'(t) for V(t) = 7t^4 - 3t^2 + 12, apply the power rule to the first term: the derivative of 7t^4 is 74t^{3} = 28t^3. For the second term, the derivative of -3t^2 is -32t^{1} = -6t. The constant term +12 has a derivative of 0, so it disappears in the derivative. A tempting distractor is choice A, which includes +12, but this is incorrect because the derivative of a constant is always zero. Always differentiate term by term, applying the power rule to each polynomial term and setting the derivative of constants to zero.

Question 15

A particle's position is s(x)=5x3+2x−9s(x)=5x^3+2x-9. What is the derivative s′(x)s'(x)?

  1. 15x2+215x^2+2 (correct answer)
  2. 15x2−215x^2-2
  3. 5x3+25x^3+2
  4. 15x3+215x^3+2
  5. 15x2+2x15x^2+2x
Explanation: This problem asks us to find the velocity function by applying basic derivative rules to the position function. We differentiate term by term: the derivative of 5x³ is 15x² (multiply by the exponent 3 and reduce the power by 1), the derivative of 2x is 2 (since x¹ becomes 1·x⁰ = 1), and the derivative of the constant -9 is 0. Therefore, s'(x) = 15x² + 2 + 0 = 15x² + 2. Choice E incorrectly keeps the x in the second term (2x instead of 2), forgetting that d/dx(2x) = 2, not 2x. The key strategy is to handle each term independently using the power rule.

Question 16

A model for a tank's water volume is V(t)=7t4−3t2+12V(t)=7t^4-3t^2+12. What is V′(t)V'(t)?

  1. 28t3−6t+1228t^3-6t+12
  2. 28t3−6t28t^3-6t (correct answer)
  3. 7t4−3t27t^4-3t^2
  4. 28t3+6t28t^3+6t
  5. 21t3−6t21t^3-6t
Explanation: This problem requires applying basic derivative rules to find the rate of change of water volume. To find V'(t), we differentiate each term separately: the derivative of 7t⁴ is 28t³ (using the power rule: bring down the 4 and reduce the exponent by 1), the derivative of -3t² is -6t, and the derivative of the constant 12 is 0. Combining these results gives V'(t) = 28t³ - 6t + 0 = 28t³ - 6t. A common error would be to include the constant term in the derivative (choice A), but constants always have a derivative of zero. Remember: when differentiating a polynomial, apply the power rule to each term and constants disappear.

Question 17

The temperature deviation is T(t)=12−7t3+2t2T(t)=12-7t^3+2t^2; compute T′(t)T'(t).

  1. −21t2+4t-21t^2+4t (correct answer)
  2. 21t2+4t21t^2+4t
  3. −21t2−4t-21t^2-4t
  4. −7t3+2t2-7t^3+2t^2
  5. −28t2+4t-28t^2+4t
Explanation: This problem assesses your understanding of the basic derivative rules for constants, sums, and differences in polynomials. To compute T'(t), differentiate each term separately using the power rule. The derivative of -7t^3 is -7 times 3t^{2}, which is -21t^2. The derivative of +2t^2 is 2 times 2t, which is +4t, and the derivative of +12 is 0, so they combine to -21t^2 + 4t. A tempting distractor like choice C changes +4t to -4t, but this fails because the positive sign remains for the quadratic term's derivative. Always remember to apply the power rule to each term independently and set the derivative of constants to zero.

Question 18

The height of water is modeled by h(t)=5t4−3t2+9h(t)=5t^4-3t^2+9; find the derivative h′(t)h'(t).

  1. 20t3−6t20t^3-6t (correct answer)
  2. 20t3+6t20t^3+6t
  3. 5t4−3t25t^4-3t^2
  4. 20t3−6t+920t^3-6t+9
  5. 25t3−6t25t^3-6t
Explanation: This problem assesses your understanding of the basic derivative rules for constants, sums, and differences in polynomials. To compute h'(t), differentiate each term separately using the power rule. The derivative of 5t^4 is 5 times 4t^{3}, which is 20t^3. The derivative of -3t^2 is -3 times 2t, which is -6t, and the derivative of +9 is 0, so they combine to 20t^3 - 6t. A tempting distractor like choice D includes the +9, but this fails because the derivative of a constant is always zero. Always remember to apply the power rule to each term independently and set the derivative of constants to zero.

Question 19

A model gives weekly profit P(x)=7x3−4x+12P(x)=7x^3-4x+12 (in dollars) for xx units sold; find P′(x)P'(x).

  1. 21x2−421x^2-4 (correct answer)
  2. 21x2+421x^2+4
  3. 7x3−4x7x^3-4x
  4. 21x2−4+1221x^2-4+12
  5. 28x2−428x^2-4
Explanation: This problem assesses your understanding of the basic derivative rules for constants, sums, and differences in polynomials. To compute P'(x), differentiate each term separately using the power rule. The derivative of 7x^3 is 7 times 3x^{2}, which is 21x^2. The derivative of -4x is -4, and the derivative of the constant +12 is 0, so they combine to 21x^2 - 4. A tempting distractor like choice D includes the +12, but this fails because the derivative of a constant is always zero. Always remember to apply the power rule to each term independently and set the derivative of constants to zero.

Question 20

A spring's potential energy is U(x)=3(5x3−2x+7)U(x)=3(5x^3-2x+7) (joules). What is U′(x)U'(x)?

  1. 45x2−645x^2-6 (correct answer)
  2. 15x2−215x^2-2
  3. 45x2−6x45x^2-6x
  4. 45x2+645x^2+6
  5. 15x2−615x^2-6
Explanation: This problem involves differentiating a function with a constant multiplier. Given U(x) = 3(5x³ - 2x + 7), we first distribute to get U(x) = 15x³ - 6x + 21, then differentiate term by term. The derivative of 15x³ is 15·3x² = 45x², the derivative of -6x is -6·1 = -6, and the derivative of 21 is 0. Therefore, U'(x) = 45x² - 6 + 0 = 45x² - 6. A mistake would be to forget to multiply the 3 through before differentiating, yielding 15x² - 6 (choice E). Always expand expressions before differentiating, then apply the power rule to each term systematically.