For f(x)=\tan(x), what is the second derivative ?
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AP Calculus AB Quiz
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.
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For f(x)=\tan(x), what is the second derivative f′′(x)?
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.
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.
For f(x)=\tan(x), what is the second derivative f′′(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.
If f(x)=sin(x2), what is the second derivative f′′(x)?
Explanation: For composite functions like f(x)=sin(x2), higher-order derivatives use the chain rule multiple times. The first derivative is f′(x)=cos(x2)⋅2x, and the second requires the product rule: f′′(x)=−sin(x2)⋅2x⋅2x+cos(x2)⋅2=2cos(x2)−4x2sin(x2), 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.
For f(x)=\sqrt{1+x}= (1+x)^{1/2}, what is f′′(x)?
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.
For y=\frac{1}{x^2}+7, what is the third derivative y(3)(x)?
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.
Let y=xe^{-x}. What is the second derivative y′′?
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.
For m(x)=x2ex, what is m′′(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.
Let g(x)=sin(2x). What is the value of g′′(x)?
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.
Let F(x)=ln(x2) for x=0. What is the value of F′′(x)?
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.
For p(x)=x21 with x=0, what is p′′(x)?
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.
A particle’s position is s(t)=t4−6t2+3t. What is the value of s′′(t)?
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.
Let f(x)=e2x. What is the third derivative f(3)(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.
If f(x)=(2x^3-1)^2, what is the second derivative f′′(x)?
Explanation: Chain rule on squares yields higher derivatives via product rule. The first is f'(x) = 12x^2 (2x^3 - 1). The second is f''(x) = 24x (2x^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.
Let g(x)=(x2+1)3. What is the second derivative g′′(x)?
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(x^2 + 1)(5x^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.
If f(x)=\frac{1}{3}x^3-4x, what is the third derivative f(3)(x)?
Explanation: Higher-order derivatives involve differentiating a function multiple times, here up to the third derivative for the polynomial ( f(x) = rac{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.
For q(x)=sin(3x), what is the second derivative q′′(x)?
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.
For y=x^2 ln(x) with x>0, what is the second derivative y′′?
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.
Let f(x)=(1-x)^4. What is the second derivative f′′(x)?
Explanation: Binomial powers differentiate with chain rule and signs. The first is f'(x) = -4(1 - x)^3. The second is f''(x) = 12(1 - x)^2, flipping positive. A common stopping error is only first derivative. Note the sign changes from the inner derivative. A transferable strategy is to apply chain rule multiple times, tracking the inner function's derivative.
Let m(x)=cos(2x). What is the third derivative m(3)(x)?
Explanation: Higher-order derivatives of cosine oscillate with sign changes. The first is m'(x) = -2 sin(2x). The second is -4 cos(2x), and third is 8 sin(2x). A common error is stopping at the second or miscounting signs. Each differentiation multiplies by -2 or 2 alternately. For periodic functions, compute derivatives step-by-step, observing the pattern for higher orders.
Let u(x)=x2lnx for x>0. What is u′′(x)?
Explanation: For u(x) = x²ln x where x > 0, we need the product rule twice. First, u'(x) = 2x ln x + x²(1/x) = 2x ln x + x. For the second derivative, differentiate each term: d/dx(2x ln x) = 2ln x + 2x(1/x) = 2ln x + 2, and d/dx(x) = 1. Therefore, u''(x) = 2ln x + 2 + 1 = 2ln x + 3. A common error is forgetting that d/dx(ln x) = 1/x when applying the product rule to the first term. Always carefully track each application of differentiation rules when products involve logarithms.
Let s(x)=cos(3x). What is s′′(x)?
Explanation: To find s''(x) for s(x) = cos(3x), we apply the chain rule twice. First, s'(x) = -sin(3x)·3 = -3sin(3x). Then s''(x) = -3cos(3x)·3 = -9cos(3x). The pattern emerges: each differentiation introduces a factor of 3 (from the chain rule) and cycles through cos → -sin → -cos → sin. A common mistake is forgetting to multiply by the derivative of the inner function (3) at each step. For trig functions with linear arguments, remember that the nth derivative introduces a factor of the argument's coefficient raised to the nth power.