Let . For which value of does ?
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AP Calculus AB Quiz
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.
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Let f(x)=4x4−2x2+5. For which value of x does f′(x)=0?
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.
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.
Let f(x)=4x4−2x2+5. For which value of x does f′(x)=0?
Explanation: First, find the derivative of f(x): f′(x)=41(4x3)−21(2x)+0=x3−x. Now, set the derivative equal to zero: x3−x=0. Factor out an x: x(x2−1)=0. This gives solutions x=0 and x2=1, so x=−1,0,1. From the given choices, x=1 is a correct answer.
If f(x)=5x3−2x2+x−9, what is f′(x)?
Explanation: To find the derivative of f(x), we apply the power rule, sum/difference rule, and constant rule to each term. The derivative of 5x3 is 15x2. The derivative of −2x2 is −4x. The derivative of x is 1. The derivative of the constant −9 is 0. Combining these gives f′(x)=15x2−4x+1.
The position of a particle moving along the x-axis is given by p(t)=t3−6t2+5 for time t≥0. What is the velocity of the particle at the instant its acceleration is zero?
Explanation: The velocity function is the first derivative of the position function: v(t)=p′(t)=3t2−12t. The acceleration function is the second derivative of the position function: a(t)=p′′(t)=6t−12. To find when the acceleration is zero, set a(t)=0: 6t−12=0, which gives t=2. To find the velocity at this time, substitute t=2 into the velocity function: v(2)=3(2)2−12(2)=3(4)−24=12−24=−12.
What is the derivative of f(x)=(2x−3)2?
Explanation: First, expand the expression for f(x): f(x)=(2x−3)(2x−3)=4x2−6x−6x+9=4x2−12x+9. Now, differentiate the expanded polynomial term by term: f'(x) = rac{d}{dx}(4x^2 - 12x + 9) = 8x - 12.
If h(x)=f(x)+g(x), where f and g are differentiable functions, and the line tangent to the graph of f at x=3 is y=2x−1 and the line tangent to the graph of g at x=3 is y=−x+7, what is h′(3)?
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 f, y=2x−1, the slope is 2. So, f′(3)=2. From the equation of the tangent line to g, y=−x+7, the slope is -1. So, g′(3)=−1. Using the sum rule for derivatives, h′(x)=f′(x)+g′(x). Therefore, h′(3)=f′(3)+g′(3)=2+(−1)=1.
Let f and g be differentiable functions. If h(x)=4f(x)−3g(x)+2, and it is known that f′(2)=5 and g′(2)=−1, what is the value of h′(2)?
Explanation: First, find the derivative of h(x) using the sum, difference, and constant multiple rules. h′(x)=4f′(x)−3g′(x)+0. Now, substitute x=2 into the expression for h′(x): h′(2)=4f′(2)−3g′(2). Using the given values, h′(2)=4(5)−3(−1)=20+3=23.
Let f(t)=9t3−2t2−6t+1. What is f′(t)?
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 x^n) = c n x^{n-1}, remember constants vanish, and maintain sign integrity in the final expression.
Let f(x)=3x2−7 and g(x)=5x+2. If h(x)=2f(x)−3g(x)+4, what is h′(x)?
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−16. Therefore, h′(x)=12x−15. Choice B incorrectly uses coefficient 6 instead of 12 for the x term. Choice C forgets the −3g(x) term in the constant. Choice D makes both errors.
Let P(x)=ax3+bx2+cx+d where a, b, c, and d are constants. If dxd[P(x)+Q(x)]=6x2−4x+7 and Q′(x)=2x2+x−3, what is P′(x)?
Explanation: When you encounter problems involving derivatives of sums, remember that the derivative of a sum equals the sum of derivatives: dxd[P(x)+Q(x)]=P′(x)+Q′(x). This fundamental property is key to solving this problem. Since you know that dxd[P(x)+Q(x)]=6x2−4x+7 and Q′(x)=2x2+x−3, you can substitute into the sum rule: P′(x)+Q′(x)=6x2−4x+7. Therefore, P′(x)+(2x2+x−3)=6x2−4x+7. Solving for P′(x): P′(x)=(6x2−4x+7)−(2x2+x−3)=6x2−4x+7−2x2−x+3=4x2−5x+10. This confirms answer choice D. Choice A (8x2−3x+10) results from incorrectly adding P′(x)+Q′(x) instead of subtracting Q′(x). Choice B (4x2−5x+4) shows correct coefficient work for the x2 and x terms but an error in the constant term (7−(−3)=10, not 4). Choice C (8x2−3x+4) combines both mistakes: adding instead of subtracting the polynomial terms AND miscalculating the constant. Always remember: when you have f′(x)+g′(x)=h′(x) and need to find one derivative, isolate it by subtracting the known derivative from both sides. Double-check your arithmetic, especially with signs when subtracting polynomials.
If F(x)=αf(x)+βg(x)+γ where α, β, γ are nonzero constants, and it's known that F′(x)=0 for all x, which of the following must be true?
Explanation: Since F′(x)=αf′(x)+βg′(x)+0=αf′(x)+βg′(x) (the derivative of constant γ is 0), and F′(x)=0 for all x, we must have αf′(x)+βg′(x)=0 for all x. 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.
If y=2x2−πx+e+31, then dxdy equals:
Explanation: Using the power rule and constant rule: dxd[2x2]=2⋅2x=22x, dxd[−πx]=−π, and dxd[e+31]=0 since both e and 31 are constants. Choice A incorrectly differentiates the constant 31. Choice B incorrectly keeps e as non-zero. Choice D uses wrong coefficient for x2 term and keeps constant e.
The function f(x)=k1g1(x)+k2g2(x)+k3g3(x) where k1=3, k2=−2, k3=5, g1′(x)=4x, g2′(x)=−x2, and g3′(x)=7. What is f′(x)?
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), you need to differentiate each term separately. Since f(x)=3g1(x)−2g2(x)+5g3(x), we have: f′(x)=3g1′(x)−2g2′(x)+5g3′(x) Substituting the given derivatives: f′(x)=3(4x)−2(−x2)+5(7) f′(x)=12x+2x2+35 This matches answer choice B. Let's examine why the other options are incorrect: Choice A (12x+2x2−35) incorrectly makes the constant term negative, likely from mishandling k3⋅g3′(x)=5⋅7=35. Choice C (12x−2x2+35) makes the x2 term negative, suggesting confusion about the double negative: k2⋅g2′(x)=(−2)⋅(−x2)=+2x2. Choice D (12x−2x2−35) combines both errors from A and C, getting both the x2 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), remember that negative times negative equals positive.
Let F(x)=3x3−4x2+5x−6. What is the derivative F′(x)?
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.
Let G(x)=−6x2+13x−15. What is the derivative G′(x)?
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.
A function is y(t)=13t2+4t+9. What is y′(t)?
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.
Let N(x)=5x4+7x3−14x+9. What is N′(x)?
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 x^n) = c n x^{n-1}, and combine using linearity.
A rate function is r(t)=−9t4+10t2+7t−3. What is r′(t)?
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 x^n) = c n x^{n-1} to each, preserve signs from differences, and ignore constants in the sum.
For S(x)=10x5−5x4+4x3−1, what is S′(x)?
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.
Let y(x)=11x9+3x5−x+8. What is y′(x)?
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 x^n) = c n x^{n-1}, and assemble using linearity.
Let f(x)=−2x4−9x3+6x2−1. What is f′(x)?
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.