AP Calculus AB Quiz: Calculating Higher Order Derivatives
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Calculating Higher Order DerivativesQuestion 1 of 20

For f(x)=sin⁡(x)+cos⁡(x)f(x)=\sin(x)+\cos(x), what is the third derivative f(3)(x)f^{(3)}(x)?

f^{(3)}(x)=-Ìos(x)+ sin(x)
f^{(3)}(x)=Ìos(x)+ sin(x)
f^{(3)}(x)=- sin(x)-Ìos(x)
f^{(3)}(x)= sin(x)-Ìos(x)
f^{(3)}(x)=Ìos(x)- sin(x)
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AP Calculus AB Quiz

AP Calculus AB Quiz: Calculating Higher Order Derivatives

Practice Calculating Higher Order Derivatives 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 Calculating Higher Order Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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

For f(x)=sin⁡(x)+cos⁡(x)f(x)=\sin(x)+\cos(x), what is the third derivative f(3)(x)f^{(3)}(x)?

  1. f^{(3)}(x)=-Ìos(x)+ sin(x)
  2. f^{(3)}(x)=Ìos(x)+ sin(x)
  3. f^{(3)}(x)=- sin(x)-Ìos(x)
  4. f^{(3)}(x)= sin(x)-Ìos(x) (correct answer)
  5. f^{(3)}(x)=Ìos(x)- sin(x)
Explanation: Sums of sine and cosine have additive derivatives. The first is f'(x) = cos(x) - sin(x). The second is -sin(x) - cos(x), and third is sin(x) - cos(x). A common stopping error is not reaching the third. Patterns cycle every four derivatives. A transferable strategy is to differentiate term by term, observing cyclic behavior for trig functions.

Question 2

For f(x)=cos⁡(x2)f(x)=\cos(x^2), what is the second derivative f′′(x)f''(x)?

  1. f''(x)=-2 sin(x^2)-4x^2Ìos(x^2) (correct answer)
  2. f''(x)=2 sin(x^2)-4x^2Ìos(x^2)
  3. f''(x)=-2x sin(x^2)
  4. f''(x)=4x^2 sin(x^2)
  5. f''(x)=-4x^2 sin(x^2)
Explanation: Higher-order derivatives of ( f(x) = cos(x2x^2) ) require repeated chain rule use, similar to trigonometric composites. First derivative: ( f'(x) = -sin(x2x^2) cdot 2x ), second: ( f''(x) = -cos(x2x^2) cdot 2x cdot 2x - sin(x2x^2) cdot 2 = -4x^2 cos(x2x^2) - 2 sin(x2x^2) ), which is ( -2sin(x2x^2) - 4x^2 cos(x2x^2) ), matching A. Each step builds on the previous by differentiating products. A typical mistake is reversing signs in the chain rule, leading to positive terms instead. Verify by computing numerically at a point if possible. Practice by breaking down composites into inner functions for reliable higher-order results.

Question 3

Let f(x)=ln⁡(5x)f(x)=\ln(5x). What is the second derivative f′′(x)f''(x) for x>0x>0?

  1. f′′(x)=−1x2f''(x)=-\frac{1}{x^2} (correct answer)
  2. f′′(x)=1x2f''(x)=\frac{1}{x^2}
  3. f′′(x)=−5x2f''(x)=-\frac{5}{x^2}
  4. f′′(x)=1xf''(x)=\frac{1}{x}
  5. f′′(x)=−15x2f''(x)=-\frac{1}{5x^2}
Explanation: Logarithmic derivatives simplify, but higher orders need care. The first is f'(x) = 1/x. The second is f''(x) = -1/x^2, with negative sign. A common error is stopping at first or ignoring the constant in ln(5x). Domain x > 0. For logs, differentiate successively, noting inverse square patterns.

Question 4

If r(x)=ln⁡(x)+x2r(x)=\ln(x)+x^2, what is the second derivative r′′(x)r''(x) for x>0x>0?

  1. r′′(x)=2−1x2r''(x)=2-\frac{1}{x^2} (correct answer)
  2. r′′(x)=2+1x2r''(x)=2+\frac{1}{x^2}
  3. r′′(x)=1x+2xr''(x)=\frac{1}{x}+2x
  4. r′′(x)=−1x+2r''(x)=-\frac{1}{x}+2
  5. r′′(x)=−1x2+2xr''(x)=-\frac{1}{x^2}+2x
Explanation: Higher-order derivatives of logarithmic and polynomial terms require combining rules. The first is r'(x) = 1/x + 2x. The second is r''(x) = -1/x^2 + 2, with the logarithmic part contributing the negative. A common error is stopping after the first or forgetting the domain x > 0. Differentiate each part separately. For such functions, differentiate term by term repeatedly, ensuring domain restrictions are noted.

Question 5

For f(x)=tan⁡(x)f(x)=\tan(x), what is the second derivative f′′(x)f''(x)?

  1. f''(x)=2 sin(x) sec^3(x)
  2. f''(x)= sec^2(x)
  3. f''(x)=2 sec^2(x) tan(x) (correct answer)
  4. f''(x)=-2 sec^2(x) tan(x)
  5. f''(x)= sec(x) tan(x)
Explanation: Tangent's derivatives involve secant and tangent terms. The first is f'(x) = sec^2(x). The second is f''(x) = 2 sec^2(x) tan(x), using chain rule. A common stopping error is only computing the first. Express in trig identities if simplifying. A transferable strategy for trig derivatives is to use known identities and chain rule repeatedly.

Question 6

If f(x)=sin⁡(x2)f(x) = \sin(x^2), what is the second derivative f′′(x)f''(x)?

  1. f′′(x)=2cos⁡(x2)−4x2sin⁡(x2)f''(x)=2\cos(x^2)-4x^2\sin(x^2) (correct answer)
  2. f′′(x)=2cos⁡(x2)+4x2sin⁡(x2)f''(x)=2\cos(x^2)+4x^2\sin(x^2)
  3. f′′(x)=2xcos⁡(x2)f''(x)=2x\cos(x^2)
  4. f′′(x)=−4x2sin⁡(x2)f''(x)=-4x^2\sin(x^2)
  5. f′′(x)=4xsin⁡(x2)f''(x)=4x\sin(x^2)
Explanation: For composite functions like f(x)=sin⁡(x2)f(x) = \sin(x^2), higher-order derivatives use the chain rule multiple times. The first derivative is f′(x)=cos⁡(x2)⋅2xf'(x) = \cos(x^2) \cdot 2x, and the second requires the product rule: f′′(x)=−sin⁡(x2)⋅2x⋅2x+cos⁡(x2)⋅2=2cos⁡(x2)−4x2sin⁡(x2)f''(x) = -\sin(x^2) \cdot 2x \cdot 2x + \cos(x^2) \cdot 2 = 2\cos(x^2) - 4x^2 \sin(x^2), matching choice A. Successive application combines chain and product rules carefully. A common error is stopping after the first chain rule application and ignoring the product in the second derivative. Double-check by expanding terms fully before simplifying. Always identify inner and outer functions to apply rules systematically for any order.

Question 7

For f(x)=1+x=(1+x)1/2f(x)=\sqrt{1+x}= (1+x)^{1/2}, what is f′′(x)f''(x)?

  1. f′′(x)=−14(1+x)−3/2f''(x)=-\frac{1}{4}(1+x)^{-3/2} (correct answer)
  2. f′′(x)=14(1+x)−3/2f''(x)=\frac{1}{4}(1+x)^{-3/2}
  3. f′′(x)=−12(1+x)−1/2f''(x)=-\frac{1}{2}(1+x)^{-1/2}
  4. f′′(x)=−34(1+x)−5/2f''(x)=-\frac{3}{4}(1+x)^{-5/2}
  5. f′′(x)=12(1+x)−1/2f''(x)=\frac{1}{2}(1+x)^{-1/2}
Explanation: Binomial roots differentiate with fractional exponents. The first is f'(x) = (1/2)(1 + x)^{-1/2}. The second is f''(x) = - (1/4)(1 + x)^{-3/2}, reducing exponent. A common error is stopping early or fraction mistakes. Chain rule applies each time. For roots, convert to exponents and differentiate iteratively, updating coefficients.

Question 8

For y=1x2+7y=\frac{1}{x^2}+7, what is the third derivative y(3)(x)y^{(3)}(x)?

  1. y(3)(x)=24x5y^{(3)}(x)=\frac{24}{x^5}
  2. y(3)(x)=−24x5y^{(3)}(x)=-\frac{24}{x^5} (correct answer)
  3. y(3)(x)=6x4y^{(3)}(x)=\frac{6}{x^4}
  4. y(3)(x)=−6x4y^{(3)}(x)=-\frac{6}{x^4}
  5. y(3)(x)=−12x5y^{(3)}(x)=-\frac{12}{x^5}
Explanation: Rational functions' higher derivatives involve increasing negative exponents. The first is y' = -2/x^3. The second is 6/x^4, and third is -24/x^5. A common error is stopping at second or sign errors. Constants like 7 vanish early. For inverses, use power rule repeatedly, alternating signs with each differentiation.

Question 9

Let y=xe^{-x}. What is the second derivative y′′y''?

  1. y′′=e−x(x−2)y''=e^{-x}(x-2) (correct answer)
  2. y′′=e−x(x+2)y''=e^{-x}(x+2)
  3. y′′=−e−x(x−2)y''=-e^{-x}(x-2)
  4. y′′=e−x(1−x)y''=e^{-x}(1-x)
  5. y′′=−e−x(1−x)y''=-e^{-x}(1-x)
Explanation: Product of x and exponential uses product rule. The first is y' = e^{-x}(1 - x). The second is y'' = e^{-x}(x - 2), factoring properly. A common error is stopping at first or sign errors in exponential. Each differentiation involves both product and chain. For such products, apply rules successively, factoring the exponential out.

Question 10

For m(x)=x2exm(x)=x^2e^x, what is m′′(x)m''(x)?

  1. (x2+2x)ex(x^2+2x)e^x
  2. (x2+4x+2)ex(x^2+4x+2)e^x (correct answer)
  3. (x2+2)ex(x^2+2)e^x
  4. (2x+1)ex(2x+1)e^x
  5. (x2+4x)ex(x^2+4x)e^x
Explanation: Finding m''(x) for m(x) = x²eˣ requires the product rule applied twice. First, m'(x) = 2xeˣ + x²eˣ = (2x + x²)eˣ = (x² + 2x)eˣ. For the second derivative, we apply the product rule to (x² + 2x)eˣ: m''(x) = (2x + 2)eˣ + (x² + 2x)eˣ = [(2x + 2) + (x² + 2x)]eˣ = (x² + 4x + 2)eˣ. A common mistake is losing track of terms when applying the product rule multiple times. When differentiating products involving eˣ, factor out eˣ after each step to simplify the algebra and clearly see the polynomial coefficient pattern.

Question 11

Let g(x)=sin⁡(2x)g(x)=\sin(2x). What is the value of g′′(x)g''(x)?

  1. 2cos⁡(2x)2\cos(2x)
  2. −4sin⁡(2x)-4\sin(2x) (correct answer)
  3. 4sin⁡(2x)4\sin(2x)
  4. −2sin⁡(2x)-2\sin(2x)
  5. 4cos⁡(2x)4\cos(2x)
Explanation: To find g''(x) for g(x) = sin(2x), we use the chain rule twice. First, g'(x) = cos(2x) · 2 = 2cos(2x), where we multiply by the derivative of the inner function 2x. Applying the chain rule again, g''(x) = -sin(2x) · 2 · 2 = -4sin(2x), since the derivative of cos is -sin and we must again multiply by 2 from the chain rule. A common error is forgetting to apply the chain rule consistently or losing track of the coefficient. When differentiating composite trigonometric functions multiple times, carefully track both the trigonometric derivatives (sin → cos → -sin → -cos) and the chain rule factors from the inner function.

Question 12

Let F(x)=ln⁡(x2)F(x)=\ln(x^2) for x≠0x\ne0. What is the value of F′′(x)F''(x)?

  1. 2x\dfrac{2}{x}
  2. −2x2-\dfrac{2}{x^2} (correct answer)
  3. 2x2\dfrac{2}{x^2}
  4. −1x2-\dfrac{1}{x^2}
  5. 1x2\dfrac{1}{x^2}
Explanation: For F(x) = ln(x²), we can use the chain rule or first simplify using logarithm properties: ln(x²) = 2ln|x|. Taking the simpler approach, F'(x) = 2(1/x) = 2/x for x ≠ 0. Then F''(x) = 2(-1/x²) = -2/x². Alternatively, using the chain rule directly: F'(x) = (1/x²)(2x) = 2/x, and F''(x) = -2/x². A common error is forgetting the negative sign when differentiating 1/x or misapplying the chain rule. When dealing with logarithmic functions, simplifying using log properties before differentiating often makes the calculation clearer and reduces errors.

Question 13

For p(x)=1x2p(x)=\dfrac{1}{x^2} with x≠0x\ne0, what is p′′(x)p''(x)?

  1. 2x3\dfrac{2}{x^3}
  2. 6x4\dfrac{6}{x^4} (correct answer)
  3. −6x4-\dfrac{6}{x^4}
  4. 2x4\dfrac{2}{x^4}
  5. −2x3-\dfrac{2}{x^3}
Explanation: To find p''(x) for p(x) = 1/x² = x⁻², we rewrite using negative exponents and apply the power rule twice. First, p'(x) = -2x⁻³ = -2/x³. Then, p''(x) = -2(-3)x⁻⁴ = 6x⁻⁴ = 6/x⁴. The key is recognizing that differentiating x⁻ⁿ gives -nx⁻ⁿ⁻¹, so negative exponents become more negative with each differentiation. A common error is mishandling the signs or incorrectly applying the power rule to negative exponents. When working with rational functions like 1/xⁿ, converting to negative exponents (x⁻ⁿ) makes differentiation more straightforward and helps avoid quotient rule complications.

Question 14

A particle's position is s(t)=t4−6t2+3ts(t)=t^4-6t^2+3t. What is the value of s′′(t)s''(t)?

  1. 12t2−1212t^2-12 (correct answer)
  2. 4t3−12t+34t^3-12t+3
  3. 12t2−612t^2-6
  4. 12t3−12t12t^3-12t
  5. 12−12t212-12t^2
Explanation: To find the second derivative s''(t) of s(t) = t⁴ - 6t² + 3t, we must differentiate twice. First, s'(t) = 4t³ - 12t + 3 using the power rule on each term. Then, differentiating again gives s''(t) = 12t² - 12 + 0 = 12t² - 12. A common error is stopping after the first derivative or incorrectly applying the power rule during the second differentiation. Notice how the constant term 3t becomes 3 after the first derivative, then 0 after the second. When finding higher-order derivatives, systematically apply differentiation rules to each term, tracking how powers decrease with each application.

Question 15

Let f(x)=e2xf(x)=e^{2x}. What is the third derivative f(3)(x)f^{(3)}(x)?

  1. f(3)(x)=8e2xf^{(3)}(x)=8e^{2x} (correct answer)
  2. f(3)(x)=6e2xf^{(3)}(x)=6e^{2x}
  3. f(3)(x)=4e2xf^{(3)}(x)=4e^{2x}
  4. f(3)(x)=2e2xf^{(3)}(x)=2e^{2x}
  5. f(3)(x)=8exf^{(3)}(x)=8e^{x}
Explanation: Exponential functions differentiate easily, with higher orders scaling by the base. The first is f'(x) = 2e^{2x}. The second is 4e^{2x}, and third is 8e^{2x}. A common stopping error is not differentiating enough times. Each step multiplies by 2. A transferable strategy is to recognize the pattern where derivatives are multiples of the original.

Question 16

If f(x)=(2x^3-1)^2, what is the second derivative f′′(x)f''(x)?

  1. f′′(x)=24x(2x3−1)+72x4f''(x)=24x(2x^3-1)+72x^4 (correct answer)
  2. f′′(x)=24x(2x3−1)f''(x)=24x(2x^3-1)
  3. f′′(x)=12x2(2x3−1)f''(x)=12x^2(2x^3-1)
  4. f′′(x)=24x(2x3−1)−72x4f''(x)=24x(2x^3-1)-72x^4
  5. f′′(x)=48x2(2x3−1)f''(x)=48x^2(2x^3-1)
Explanation: Chain rule on squares yields higher derivatives via product rule. The first is f'(x) = 12x^2 (2x32x^3 - 1). The second is f''(x) = 24x (2x32x^3 - 1) + 72x^4, expanding properly. A common stopping error is not applying product rule for the second. Simplify by factoring if possible. A transferable strategy is to treat as composition and differentiate step-by-step, verifying with expansion.

Question 17

Let g(x)=(x2+1)3g(x)=(x^2+1)^3. What is the second derivative g′′(x)g''(x)?

  1. g′′(x)=6(x2+1)2g''(x)=6(x^2+1)^2
  2. g′′(x)=6(x2+1)(5x2+1)g''(x)=6(x^2+1)(5x^2+1) (correct answer)
  3. g′′(x)=12x(x2+1)2g''(x)=12x(x^2+1)^2
  4. g′′(x)=6(x2+1)(3x2+1)g''(x)=6(x^2+1)(3x^2+1)
  5. g′′(x)=12x(x2+1)(3x2+1)g''(x)=12x(x^2+1)(3x^2+1)
Explanation: Higher-order derivatives of composite functions use the chain rule repeatedly, here for a power function. The first derivative is g'(x) = 6x(x^2 + 1)^2, applying the chain rule. The second derivative requires the product rule: g''(x) = 6(x2x^2 + 1)(5x25x^2 + 1), factoring for simplicity. A common error is stopping after the first derivative or misapplying the product rule. Factor expressions to verify accuracy. For higher-order derivatives of composites, apply the chain and product rules step-by-step, simplifying at each stage for clarity.

Question 18

If f(x)=13x3−4xf(x)=\frac{1}{3}x^3-4x, what is the third derivative f(3)(x)f^{(3)}(x)?

  1. f(3)(x)=2f^{(3)}(x)=2 (correct answer)
  2. f(3)(x)=0f^{(3)}(x)=0
  3. f(3)(x)=6xf^{(3)}(x)=6x
  4. f(3)(x)=x2−4f^{(3)}(x)=x^2-4
  5. f(3)(x)=6f^{(3)}(x)=6
Explanation: Higher-order derivatives involve differentiating a function multiple times, here up to the third derivative for the polynomial ( f(x) = \frac{1}{3}x^3 - 4x ). Start by finding the first derivative: ( f'(x) = x^2 - 4 ), then the second: ( f''(x) = 2x ), and finally the third: ( f'''(x) = 2 ), which matches choice A. Each differentiation reduces the degree of the polynomial by one, leading to a constant for the third derivative. A common error is stopping at the second derivative and mistaking ( 2x ) for the third, but successive differentiation must continue until the required order. Another pitfall is misapplying the power rule, like forgetting to divide by the exponent properly. To master higher-order derivatives, always verify each step by differentiating term by term and checking the order required.

Question 19

For q(x)=sin⁡(3x)q(x)=\sin(3x), what is the second derivative q′′(x)q''(x)?

  1. q''(x)=-9 sin(3x) (correct answer)
  2. q''(x)=-3 sin(3x)
  3. q′′(x)=9Iˋos(3x)q''(x)=9Ìos(3x)
  4. q′′(x)=−9Iˋos(3x)q''(x)=-9Ìos(3x)
  5. q′′(x)=3Iˋos(3x)q''(x)=3Ìos(3x)
Explanation: Trigonometric functions have cyclic higher-order derivatives, here for sin(3x). The first is q'(x) = 3 cos(3x). The second is q''(x) = -9 sin(3x), applying the chain rule twice. A common stopping error is only finding the first derivative. Note the coefficient multiplies by -3 each time. A transferable strategy for trig functions is to apply the chain rule iteratively, tracking amplitude changes.

Question 20

For y=x^2 ln(x) with x>0x>0, what is the second derivative y′′y''?

  1. y''=2 ln(x)+3 (correct answer)
  2. y''=2 ln(x)+1
  3. y''=2x ln(x)+x
  4. y''=2 ln(x)-3
  5. y''=rac{2}{x}+3
Explanation: Products of polynomials and logs require product rule for derivatives. The first is y' = 2x ln(x) + x. The second is y'' = 2 ln(x) + 3, simplifying terms. A common error is stopping at the first or forgetting the 1/x from ln. Domain x > 0 is key. For such functions, use product rule successively, combining like terms.