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  2. AP Calculus AB
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AP Calculus AB Flashcards: Connecting A Function And Its Derivatives

Study Connecting A Function And Its Derivatives in AP Calculus AB 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 Connecting A Function And Its Derivatives, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

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 AB Flashcards: Connecting A Function And Its Derivatives

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QUESTION

Find f′′(x)f''(x)f′′(x) for f(x)=x3−5xf(x) = x^3 - 5xf(x)=x3−5x.

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ANSWER

f′′(x)=6xf''(x) = 6xf′′(x)=6x. First find f′(x)=3x2−5f'(x) = 3x^2 - 5f′(x)=3x2−5, then differentiate.

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Flashcard 1: Find f′′(x)f''(x)f′′(x) for f(x)=x3−5xf(x) = x^3 - 5xf(x)=x3−5x.

Answer: f′′(x)=6xf''(x) = 6xf′′(x)=6x. First find f′(x)=3x2−5f'(x) = 3x^2 - 5f′(x)=3x2−5, then differentiate.

Flashcard 2: What is the derivative of a constant function f(x)=cf(x) = cf(x)=c?

Answer: The derivative is 000. Constants have zero rate of change.

Flashcard 3: What indicates a local maximum in terms of derivatives?

Answer: f′(x)=0f'(x) = 0f′(x)=0 and f′′(x)<0f''(x) < 0f′′(x)<0. Critical point test: zero slope, negative concavity.

Flashcard 4: What is f′′(x)f''(x)f′′(x) for f(x)=ex+x2f(x) = \text{e}^x + x^2f(x)=ex+x2?

Answer: f′′(x)=ex+2f''(x) = \text{e}^x + 2f′′(x)=ex+2. Sum rule: derivatives of exe^xex and x2x^2x2 separately.

Flashcard 5: What is the first derivative of f(x)=1x2f(x) = \frac{1}{x^2}f(x)=x21​?

Answer: f′(x)=−2x3f'(x) = -\frac{2}{x^3}f′(x)=−x32​. Rewrite as x−2x^{-2}x−2, then apply power rule.

Flashcard 6: State the relationship between concavity and the second derivative.

Answer: Concave up if f′′(x)>0f''(x) > 0f′′(x)>0, concave down if f′′(x)<0f''(x) < 0f′′(x)<0. Second derivative test for concavity direction.

Flashcard 7: Find the second derivative: f(x)=2x4−3x3+xf(x) = 2x^4 - 3x^3 + xf(x)=2x4−3x3+x.

Answer: f′′(x)=24x2−18xf''(x) = 24x^2 - 18xf′′(x)=24x2−18x. First find f′(x)=8x3−9x2+1f'(x) = 8x^3 - 9x^2 + 1f′(x)=8x3−9x2+1, then differentiate.

Flashcard 8: What is the derivative of f(x)=exf(x) = e^xf(x)=ex?

Answer: The derivative is f′(x)=exf'(x) = e^xf′(x)=ex. The exponential function is its own derivative.

Flashcard 9: What is f′′(x)f''(x)f′′(x) if f(x)=x5−3x2+1f(x) = x^5 - 3x^2 + 1f(x)=x5−3x2+1?

Answer: f′′(x)=20x3−6f''(x) = 20x^3 - 6f′′(x)=20x3−6. First find f′(x)=5x4−6xf'(x) = 5x^4 - 6xf′(x)=5x4−6x, then differentiate.

Flashcard 10: What is the second derivative of f(x)=sin(x)f(x) = \text{sin}(x)f(x)=sin(x)?

Answer: f′′(x)=−sin(x)f''(x) = -\text{sin}(x)f′′(x)=−sin(x). First derivative is cos⁡(x)\cos(x)cos(x), then −sin⁡(x)-\sin(x)−sin(x).

Flashcard 11: Determine f′(x)f'(x)f′(x) for f(x)=cos(x)f(x) = \text{cos}(x)f(x)=cos(x).

Answer: f′(x)=−sin(x)f'(x) = -\text{sin}(x)f′(x)=−sin(x). Standard trigonometric derivative formula.

Flashcard 12: If f′′(x)=0f''(x) = 0f′′(x)=0, what can be inferred about f(x)f(x)f(x)?

Answer: f(x)f(x)f(x) may have a point of inflection. Second derivative test for inflection points.

Flashcard 13: Evaluate the second derivative: f(x)=6x−4x2f(x) = 6x - 4x^2f(x)=6x−4x2.

Answer: f′′(x)=−8f''(x) = -8f′′(x)=−8. First find f′(x)=6−8xf'(x) = 6 - 8xf′(x)=6−8x, then differentiate.

Flashcard 14: Determine f′′(x)f''(x)f′′(x) for f(x)=3x4+x2f(x) = 3x^4 + x^2f(x)=3x4+x2.

Answer: f′′(x)=36x2+2f''(x) = 36x^2 + 2f′′(x)=36x2+2. First find f′(x)=12x3+2xf'(x) = 12x^3 + 2xf′(x)=12x3+2x, then differentiate.

Flashcard 15: If f′(x)>0f'(x) > 0f′(x)>0 for all xxx, what can you say about f(x)f(x)f(x)?

Answer: f(x)f(x)f(x) is increasing. Positive derivative indicates upward slope.

Flashcard 16: State the Power Rule for differentiation.

Answer: If f(x)=xnf(x) = x^nf(x)=xn, then f′(x)=nxn−1f'(x) = nx^{n-1}f′(x)=nxn−1. Bring down the exponent and reduce power by 1.

Flashcard 17: Determine the second derivative: f(x)=4x2+3xf(x) = 4x^2 + 3xf(x)=4x2+3x.

Answer: f′′(x)=8f''(x) = 8f′′(x)=8. First derivative is 8x+38x + 38x+3, then derivative is 888.

Flashcard 18: What is the second derivative of f(x)=x3f(x) = x^3f(x)=x3?

Answer: f′′(x)=6xf''(x) = 6xf′′(x)=6x. Apply power rule twice: 3x23x^23x2 then 6x6x6x.

Flashcard 19: If f′(x)<0f'(x) < 0f′(x)<0 for all xxx, what is true about f(x)f(x)f(x)?

Answer: f(x)f(x)f(x) is decreasing. Negative derivative indicates downward slope.

Flashcard 20: Evaluate f′(x)f'(x)f′(x) for f(x)=tan(x)f(x) = \text{tan}(x)f(x)=tan(x).

Answer: f′(x)=sec2(x)f'(x) = \text{sec}^2(x)f′(x)=sec2(x). Standard trigonometric derivative formula.

Flashcard 21: Determine the derivative of f(x)=1x3f(x) = \frac{1}{x^3}f(x)=x31​.

Answer: f′(x)=−3x4f'(x) = -\frac{3}{x^4}f′(x)=−x43​. Rewrite as x−3x^{-3}x−3 and apply power rule.

Flashcard 22: Evaluate f′(x)f'(x)f′(x) for f(x)=5x3−4x2+2x−7f(x) = 5x^3 - 4x^2 + 2x - 7f(x)=5x3−4x2+2x−7.

Answer: f′(x)=15x2−8x+2f'(x) = 15x^2 - 8x + 2f′(x)=15x2−8x+2. Apply power rule to each term separately.

Flashcard 23: Evaluate f′(x)f'(x)f′(x) for f(x)=5x3−4x2+2x−7f(x) = 5x^3 - 4x^2 + 2x - 7f(x)=5x3−4x2+2x−7.

Answer: f′(x)=15x2−8x+2f'(x) = 15x^2 - 8x + 2f′(x)=15x2−8x+2. Apply power rule to each term separately.

Flashcard 24: Determine the derivative of f(x)=1x3f(x) = \frac{1}{x^3}f(x)=x31​.

Answer: f′(x)=−3x4f'(x) = -\frac{3}{x^4}f′(x)=−x43​. Rewrite as x−3x^{-3}x−3 and apply power rule.

Flashcard 25: Determine f′′(x)f''(x)f′′(x) for f(x)=3x4+x2f(x) = 3x^4 + x^2f(x)=3x4+x2.

Answer: f′′(x)=36x2+2f''(x) = 36x^2 + 2f′′(x)=36x2+2. First find f′(x)=12x3+2xf'(x) = 12x^3 + 2xf′(x)=12x3+2x, then differentiate.

Flashcard 26: Evaluate the second derivative: f(x)=6x−4x2f(x) = 6x - 4x^2f(x)=6x−4x2.

Answer: f′′(x)=−8f''(x) = -8f′′(x)=−8. First find f′(x)=6−8xf'(x) = 6 - 8xf′(x)=6−8x, then differentiate.

Flashcard 27: Evaluate f′(x)f'(x)f′(x) for f(x)=tan(x)f(x) = \text{tan}(x)f(x)=tan(x).

Answer: f′(x)=sec2(x)f'(x) = \text{sec}^2(x)f′(x)=sec2(x). Standard trigonometric derivative formula.

Flashcard 28: If f′(x)<0f'(x) < 0f′(x)<0 for all xxx, what is true about f(x)f(x)f(x)?

Answer: f(x)f(x)f(x) is decreasing. Negative derivative indicates downward slope.

Flashcard 29: What is the second derivative of f(x)=x3f(x) = x^3f(x)=x3?

Answer: f′′(x)=6xf''(x) = 6xf′′(x)=6x. Apply power rule twice: 3x23x^23x2 then 6x6x6x.

Flashcard 30: State the Power Rule for differentiation.

Answer: If f(x)=xnf(x) = x^nf(x)=xn, then f′(x)=nxn−1f'(x) = nx^{n-1}f′(x)=nxn−1. Bring down the exponent and reduce power by 1.