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  2. AP Calculus AB
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AP Calculus AB Flashcards: Sketching Graphs Of Functions And Derivatives

Study Sketching Graphs Of Functions And 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 Sketching Graphs Of Functions And 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: Sketching Graphs Of Functions And Derivatives

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QUESTION

What does f′(x)=0f'(x) = 0f′(x)=0 indicate about f(x)f(x)f(x)?

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ANSWER

Potential extrema. Horizontal tangent lines occur at critical points.

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Flashcard 1: What does f′(x)=0f'(x) = 0f′(x)=0 indicate about f(x)f(x)f(x)?

Answer: Potential extrema. Horizontal tangent lines occur at critical points.

Flashcard 2: What is the derivative of f(x)=arcsin(x)f(x) = \text{arcsin}(x)f(x)=arcsin(x)?

Answer: f′(x)=1√(1−x2)f'(x) = \frac{1}{\text{√}(1-x^2)}f′(x)=√(1−x2)1​. Inverse trig derivative with radical denominator.

Flashcard 3: State the Quotient Rule for differentiation.

Answer: ddx[uv]=u′v−uv′v2\frac{d}{dx}[\frac{u}{v}] = \frac{u'v - uv'}{v^2}dxd​[vu​]=v2u′v−uv′​. Low d-high minus high d-low over low squared.

Flashcard 4: What is the derivative of f(x)=cot(x)f(x) = \text{cot}(x)f(x)=cot(x)?

Answer: f′(x)=−csc2(x)f'(x) = -\text{csc}^2(x)f′(x)=−csc2(x). Derivative of cotangent is negative cosecant squared.

Flashcard 5: Differentiate f(x)=1xf(x) = \frac{1}{x}f(x)=x1​.

Answer: f′(x)=−1x2f'(x) = -\frac{1}{x^2}f′(x)=−x21​. Rewrite as x−1x^{-1}x−1 and apply power rule.

Flashcard 6: Identify the concavity of f(x)=x4f(x) = x^4f(x)=x4.

Answer: Concave up for all xxx. f′′(x)=12x2≥0f''(x) = 12x^2 \geq 0f′′(x)=12x2≥0 for all real x.

Flashcard 7: Identify the critical points of f(x)=x3−3xf(x) = x^3 - 3xf(x)=x3−3x.

Answer: x=0,x=±√33x = 0, x = \text{±}\frac{\text{√}3}{3}x=0,x=±3√3​. Set f′(x)=3x2−3=0f'(x) = 3x^2 - 3 = 0f′(x)=3x2−3=0 and solve for x.

Flashcard 8: Find f′(x)f'(x)f′(x) for f(x)=ln(x)f(x) = \text{ln}(x)f(x)=ln(x).

Answer: f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1​. The natural logarithm's derivative is 1x\frac{1}{x}x1​.

Flashcard 9: What is the inverse function derivative formula?

Answer: [f−1]′(x)=1f′(f−1(x))[f^{-1}]'(x) = \frac{1}{f'(f^{-1}(x))}[f−1]′(x)=f′(f−1(x))1​. Reciprocal of derivative at corresponding point.

Flashcard 10: What is f′(x)f'(x)f′(x) for f(x)=sin(x)f(x) = \text{sin}(x)f(x)=sin(x)?

Answer: f′(x)=cos(x)f'(x) = \text{cos}(x)f′(x)=cos(x). Derivative of sine is cosine.

Flashcard 11: Which test identifies concavity?

Answer: Second Derivative Test. Examines sign of f′′(x)f''(x)f′′(x) to determine concavity.

Flashcard 12: What is f′(x)f'(x)f′(x) for f(x)=x5f(x) = x^5f(x)=x5?

Answer: f′(x)=5x4f'(x) = 5x^4f′(x)=5x4. Power rule: bring down 5, subtract 1 from exponent.

Flashcard 13: State the Chain Rule for differentiation.

Answer: ddx[f(g(x))]=f′(g(x))g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)dxd​[f(g(x))]=f′(g(x))g′(x). Differentiate outer function times inner derivative.

Flashcard 14: State the Power Rule for differentiation.

Answer: ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}dxd​xn=nxn−1. Multiply by exponent, reduce exponent by 1.

Flashcard 15: What is the second derivative of f(x)=3x4f(x) = 3x^4f(x)=3x4?

Answer: f′′(x)=36x2f''(x) = 36x^2f′′(x)=36x2. Apply power rule twice: f′(x)=12x3f'(x) = 12x^3f′(x)=12x3, then again.

Flashcard 16: Calculate 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). Derivative of cosine is negative sine.

Flashcard 17: What is the derivative of f(x)=exf(x) = e^xf(x)=ex?

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

Flashcard 18: What is the Product Rule for differentiation?

Answer: ddx[uv]=u′v+uv′\frac{d}{dx}[uv] = u'v + uv'dxd​[uv]=u′v+uv′. Sum of each function times the other's derivative.

Flashcard 19: Differentiate f(x)=7x3−2xf(x) = 7x^3 - 2xf(x)=7x3−2x.

Answer: f′(x)=21x2−2f'(x) = 21x^2 - 2f′(x)=21x2−2. Apply power rule to each term separately.

Flashcard 20: What is the derivative of a constant ccc?

Answer: 000. Constants have zero rate of change.

Flashcard 21: Find the critical points of f(x)=x2−6x+8f(x) = x^2 - 6x + 8f(x)=x2−6x+8.

Answer: x=3x = 3x=3. Set f′(x)=2x−6=0f'(x) = 2x - 6 = 0f′(x)=2x−6=0 and solve.

Flashcard 22: Find f′(x)f'(x)f′(x) for f(x)=arccot(x)f(x) = \text{arccot}(x)f(x)=arccot(x).

Answer: f′(x)=−11+x2f'(x) = -\frac{1}{1+x^2}f′(x)=−1+x21​. Negative of arctangent derivative.

Flashcard 23: What is f′(x)f'(x)f′(x) for f(x)=arctan(x)f(x) = \text{arctan}(x)f(x)=arctan(x)?

Answer: f′(x)=11+x2f'(x) = \frac{1}{1+x^2}f′(x)=1+x21​. Inverse tangent derivative with squared denominator.

Flashcard 24: Calculate f′(x)f'(x)f′(x) for f(x)=arccos(x)f(x) = \text{arccos}(x)f(x)=arccos(x).

Answer: f′(x)=−1√(1−x2)f'(x) = -\frac{1}{\text{√}(1-x^2)}f′(x)=−√(1−x2)1​. Negative of arcsine derivative.

Flashcard 25: Find f′(x)f'(x)f′(x) for f(x)=csc(x)f(x) = \text{csc}(x)f(x)=csc(x).

Answer: f′(x)=−csc(x)cot(x)f'(x) = -\text{csc}(x)\text{cot}(x)f′(x)=−csc(x)cot(x). Derivative involves negative cotangent cosecant.

Flashcard 26: Calculate f′(x)f'(x)f′(x) for f(x)=12x4f(x) = \frac{1}{2}x^4f(x)=21​x4.

Answer: f′(x)=2x3f'(x) = 2x^3f′(x)=2x3. Apply power rule: 4⋅12⋅x3=2x34 \cdot \frac{1}{2} \cdot x^3 = 2x^34⋅21​⋅x3=2x3.

Flashcard 27: What is the derivative of f(x)=x2f(x) = x^2f(x)=x2?

Answer: f′(x)=2xf'(x) = 2xf′(x)=2x. Power rule: bring down exponent, subtract 1.

Flashcard 28: What is the derivative of 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). Derivative of tangent is secant squared.

Flashcard 29: Which test uses f′′(x)f''(x)f′′(x) to find extrema?

Answer: Second Derivative Test. Uses second derivative to classify critical points.

Flashcard 30: Find f′(x)f'(x)f′(x) for f(x)=ln(x)f(x) = \text{ln}(x)f(x)=ln(x).

Answer: f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1​. The natural logarithm's derivative is 1x\frac{1}{x}x1​.