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AP Calculus BC Flashcards: The Product Rule

Study The Product Rule in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on The Product Rule, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: The Product Rule

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QUESTION

What is the derivative of f(x)=x3×ln⁡(x)f(x) = x^3 \times \ln(x)f(x)=x3×ln(x) using the Product Rule?

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ANSWER

3x2×ln⁡(x)+x23x^2 \times \ln(x) + x^23x2×ln(x)+x2. Use (fg)′=f′g+fg′(fg)' = f'g + fg'(fg)′=f′g+fg′ where ddx[x3]=3x2\frac{d}{dx}[x^3] = 3x^2dxd​[x3]=3x2 and ddx[ln⁡(x)]=1x\frac{d}{dx}[\ln(x)] = \frac{1}{x}dxd​[ln(x)]=x1​.

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Flashcard 1: What is the derivative of f(x)=x3×ln⁡(x)f(x) = x^3 \times \ln(x)f(x)=x3×ln(x) using the Product Rule?

Answer: 3x2×ln⁡(x)+x23x^2 \times \ln(x) + x^23x2×ln(x)+x2. Use (fg)′=f′g+fg′(fg)' = f'g + fg'(fg)′=f′g+fg′ where ddx[x3]=3x2\frac{d}{dx}[x^3] = 3x^2dxd​[x3]=3x2 and ddx[ln⁡(x)]=1x\frac{d}{dx}[\ln(x)] = \frac{1}{x}dxd​[ln(x)]=x1​.

Flashcard 2: Apply the Product Rule to find f′(x)f'(x)f′(x) for f(x)=(x+1)×exf(x) = (x + 1) \times \text{e}^xf(x)=(x+1)×ex.

Answer: ex+(x+1)×exe^x + (x + 1) \times e^xex+(x+1)×ex. Factor out exe^xex to get ex(1+x+1)=ex(2+x)e^x(1 + x + 1) = e^x(2 + x)ex(1+x+1)=ex(2+x).

Flashcard 3: Identify the derivative of f(x)=ex×sin(x)f(x) = e^x \times \text{sin}(x)f(x)=ex×sin(x) using the Product Rule.

Answer: ex×sin(x)+ex×cos(x)e^x \times \text{sin}(x) + e^x \times \text{cos}(x)ex×sin(x)+ex×cos(x). Use (fg)′=f′g+fg′(fg)' = f'g + fg'(fg)′=f′g+fg′ with derivatives of exe^xex and sin⁡(x)\sin(x)sin(x).

Flashcard 4: Derive the function f(x)=x2×e2xf(x) = x^2 \times \text{e}^{2x}f(x)=x2×e2x using the Product Rule.

Answer: 2x×e2x+2x2×e2x2x \times e^{2x} + 2x^2 \times e^{2x}2x×e2x+2x2×e2x. Factor out 2xe2x2x e^{2x}2xe2x to get 2xe2x(1+x)2x e^{2x}(1 + x)2xe2x(1+x).

Flashcard 5: What is the derivative of f(x)=x3×sin(x)f(x) = x^3 \times \text{sin}(x)f(x)=x3×sin(x) using the Product Rule?

Answer: 3x2×sin(x)+x3×cos(x)3x^2 \times \text{sin}(x) + x^3 \times \text{cos}(x)3x2×sin(x)+x3×cos(x). Factor out x2x^2x2 to get x2(3sin⁡(x)+xcos⁡(x))x^2(3\sin(x) + x\cos(x))x2(3sin(x)+xcos(x)).

Flashcard 6: Determine the derivative of y=(3x)×exy = (3x) \times \text{e}^xy=(3x)×ex using the Product Rule.

Answer: 3ex+3x×ex3e^x + 3x \times e^x3ex+3x×ex. Factor out 3ex3e^x3ex to simplify to 3ex(1+x)3e^x(1 + x)3ex(1+x).

Flashcard 7: Calculate the derivative using the Product Rule for y=(1−x)×(2+x)y = (1 - x) \times (2 + x)y=(1−x)×(2+x).

Answer: −(2+x)+(1−x)-(2 + x) + (1 - x)−(2+x)+(1−x). Expand and simplify to get −1−2x-1 - 2x−1−2x.

Flashcard 8: What is the derivative of f(x)=x×tan(x)f(x) = x \times \text{tan}(x)f(x)=x×tan(x) using the Product Rule?

Answer: tan(x)+x×sec2(x)\text{tan}(x) + x \times \text{sec}^2(x)tan(x)+x×sec2(x). Apply the product rule with derivatives 111 and sec⁡2(x)\sec^2(x)sec2(x).

Flashcard 9: Which option correctly applies the Product Rule to f(x)=x3×cos(x)f(x) = x^3 \times \text{cos}(x)f(x)=x3×cos(x)?

Answer: 3x2×cos(x)−x3×sin(x)3x^2 \times \text{cos}(x) - x^3 \times \text{sin}(x)3x2×cos(x)−x3×sin(x). Apply the product rule: derivative of first times second plus first times derivative of second.

Flashcard 10: Calculate the derivative using the Product Rule for f(x)=2x×sin(x)f(x) = 2x \times \text{sin}(x)f(x)=2x×sin(x).

Answer: 2×sin(x)+2x×cos(x)2 \times \text{sin}(x) + 2x \times \text{cos}(x)2×sin(x)+2x×cos(x). Factor out 222 to get 2(sin⁡(x)+xcos⁡(x))2(\sin(x) + x\cos(x))2(sin(x)+xcos(x)).

Flashcard 11: Find the derivative using the Product Rule for f(x)=x×cosh(x)f(x) = x \times \text{cosh}(x)f(x)=x×cosh(x).

Answer: cosh(x)+x×sinh(x)\text{cosh}(x) + x \times \text{sinh}(x)cosh(x)+x×sinh(x). Apply the product rule with hyperbolic function derivatives.

Flashcard 12: Derive the function f(x)=5x×e−xf(x) = 5x \times \text{e}^{-x}f(x)=5x×e−x using the Product Rule.

Answer: 5e−x−5x×e−x5e^{-x} - 5x \times e^{-x}5e−x−5x×e−x. Factor out 5e−x5e^{-x}5e−x to get 5e−x(1−x)5e^{-x}(1 - x)5e−x(1−x).

Flashcard 13: Calculate the derivative of f(x)=(x+1)×(x−2)f(x) = (x + 1) \times (x - 2)f(x)=(x+1)×(x−2) using the Product Rule.

Answer: (x−2)+(x+1)(x - 2) + (x + 1)(x−2)+(x+1). Simplifies to 2x−12x - 12x−1 when expanded and combined.

Flashcard 14: Apply the Product Rule to f(x)=sin(x)×cos(x)f(x) = \text{sin}(x) \times \text{cos}(x)f(x)=sin(x)×cos(x).

Answer: cos2(x)−sin2(x)\text{cos}^2(x) - \text{sin}^2(x)cos2(x)−sin2(x). This simplifies to cos⁡(2x)\cos(2x)cos(2x) using the double angle identity.

Flashcard 15: Identify the derivative of f(x)=x3×exf(x) = x^3 \times \text{e}^xf(x)=x3×ex using the Product Rule.

Answer: 3x2×ex+x3×ex3x^2 \times e^x + x^3 \times e^x3x2×ex+x3×ex. Factor out x2exx^2 e^xx2ex to get x2ex(3+x)x^2 e^x(3 + x)x2ex(3+x).

Flashcard 16: What is the derivative of f(x)=x×sin(x)f(x) = x \times \text{sin}(x)f(x)=x×sin(x) using the Product Rule?

Answer: sin(x)+x×cos(x)\text{sin}(x) + x \times \text{cos}(x)sin(x)+x×cos(x). Apply the product rule with derivatives 111 and cos⁡(x)\cos(x)cos(x).

Flashcard 17: Compute the derivative of f(x)=x×cosh(x)f(x) = x \times \text{cosh}(x)f(x)=x×cosh(x) using the Product Rule.

Answer: cosh(x)+x×sinh(x)\text{cosh}(x) + x \times \text{sinh}(x)cosh(x)+x×sinh(x). Apply (fg)′=f′g+fg′(fg)' = f'g + fg'(fg)′=f′g+fg′ with hyperbolic function derivatives.

Flashcard 18: Using the Product Rule, what is the derivative of f(x)=x×log(x)f(x) = x \times \text{log}(x)f(x)=x×log(x)?

Answer: log(x)+1\text{log}(x) + 1log(x)+1. Use the fact that ddx[xlog⁡(x)]=log⁡(x)+1\frac{d}{dx}[x \log(x)] = \log(x) + 1dxd​[xlog(x)]=log(x)+1.

Flashcard 19: Determine the derivative of f(x)=x2×exf(x) = x^2 \times \text{e}^xf(x)=x2×ex using the Product Rule.

Answer: 2x×ex+x2×ex2x \times e^x + x^2 \times e^x2x×ex+x2×ex. Factor out xexx e^xxex to get xex(2+x)x e^x(2 + x)xex(2+x).

Flashcard 20: Compute the derivative using the Product Rule for y=x×cos(x)y = x \times \text{cos}(x)y=x×cos(x).

Answer: cos(x)−x×sin(x)\text{cos}(x) - x \times \text{sin}(x)cos(x)−x×sin(x). Apply the product rule with derivatives 111 and −sin⁡(x)-\sin(x)−sin(x).

Flashcard 21: Using the Product Rule, find the derivative for y=x2×tan(x)y = x^2 \times \text{tan}(x)y=x2×tan(x).

Answer: 2x×tan(x)+x2×sec2(x)2x \times \text{tan}(x) + x^2 \times \text{sec}^2(x)2x×tan(x)+x2×sec2(x). Factor out xxx to get x(2tan⁡(x)+xsec⁡2(x))x(2\tan(x) + x\sec^2(x))x(2tan(x)+xsec2(x)).

Flashcard 22: What is the derivative of f(x)=x4×tan(x)f(x) = x^4 \times \text{tan}(x)f(x)=x4×tan(x) using the Product Rule?

Answer: 4x3×tan(x)+x4×sec2(x)4x^3 \times \text{tan}(x) + x^4 \times \text{sec}^2(x)4x3×tan(x)+x4×sec2(x). Factor out x3x^3x3 to get x3(4tan⁡(x)+xsec⁡2(x))x^3(4\tan(x) + x\sec^2(x))x3(4tan(x)+xsec2(x)).

Flashcard 23: Calculate the derivative using the Product Rule for y=(2x)×ln(x)y = (2x) \times \text{ln}(x)y=(2x)×ln(x).

Answer: 2×ln(x)+2xx2 \times \text{ln}(x) + \frac{2x}{x}2×ln(x)+x2x​. Simplifies to 2ln⁡(x)+22\ln(x) + 22ln(x)+2 since 2xx=2\frac{2x}{x} = 2x2x​=2.

Flashcard 24: Find the derivative of f(x)=(x2+1)(x3−x)f(x) = (x^2 + 1)(x^3 - x)f(x)=(x2+1)(x3−x) using the Product Rule.

Answer: (2x)(x3−x)+(x2+1)(3x2−1)(2x)(x^3 - x) + (x^2 + 1)(3x^2 - 1)(2x)(x3−x)+(x2+1)(3x2−1). Apply product rule to each factor and combine the terms.

Flashcard 25: What is the derivative of f(x)=x2×ln(x)f(x) = x^2 \times \text{ln}(x)f(x)=x2×ln(x) using the Product Rule?

Answer: 2x×ln(x)+x2x \times \text{ln}(x) + x2x×ln(x)+x. Apply (fg)′=f′g+fg′(fg)' = f'g + fg'(fg)′=f′g+fg′ with f=x2f = x^2f=x2 and g=ln⁡(x)g = \ln(x)g=ln(x).

Flashcard 26: State the formula for the Product Rule in calculus.

Answer: (fg)′=f′g+fg′(fg)' = f'g + fg'(fg)′=f′g+fg′. The fundamental formula where each function's derivative multiplies the other function.

Flashcard 27: What is the result of applying the Product Rule to y=x×e2xy = x \times e^{2x}y=x×e2x?

Answer: e2x+2xe2xe^{2x} + 2xe^{2x}e2x+2xe2x. Factor out e2xe^{2x}e2x to get e2x(1+2x)e^{2x}(1 + 2x)e2x(1+2x) after applying the rule.

Flashcard 28: Identify the derivative of f(x)=x5×e2xf(x) = x^5 \times \text{e}^{2x}f(x)=x5×e2x using the Product Rule.

Answer: 5x4×e2x+2x5×e2x5x^4 \times e^{2x} + 2x^5 \times e^{2x}5x4×e2x+2x5×e2x. Factor out x4e2xx^4 e^{2x}x4e2x to get x4e2x(5+2x)x^4 e^{2x}(5 + 2x)x4e2x(5+2x).

Flashcard 29: Which is the correct application of the Product Rule to f(x)=3x×ln(x)f(x) = 3x \times \text{ln}(x)f(x)=3x×ln(x)?

Answer: 3×ln(x)+3xx3 \times \text{ln}(x) + \frac{3x}{x}3×ln(x)+x3x​. Simplifies to 3ln⁡(x)+33 \ln(x) + 33ln(x)+3 since 3xx=3\frac{3x}{x} = 3x3x​=3.

Flashcard 30: What is the derivative of f(x)=x2×e−xf(x) = x^2 \times \text{e}^{-x}f(x)=x2×e−x using the Product Rule?

Answer: 2x×e−x−x2×e−x2x \times e^{-x} - x^2 \times e^{-x}2x×e−x−x2×e−x. Factor out xe−xx e^{-x}xe−x to get xe−x(2−x)x e^{-x}(2 - x)xe−x(2−x).