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  2. AP Calculus BC
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AP Calculus BC Flashcards: Introducing Calculus

Study Introducing Calculus 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 Introducing Calculus, 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: Introducing Calculus

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QUESTION

State the formula for the derivative of f(x)f(x)f(x) using limits.

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ANSWER

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}f′(x)=limh→0​hf(x+h)−f(x)​. This is the formal limit definition of the derivative.

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Flashcard 1: State the formula for the derivative of f(x)f(x)f(x) using limits.

Answer: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}f′(x)=limh→0​hf(x+h)−f(x)​. This is the formal limit definition of the derivative.

Flashcard 2: Identify the derivative of f(x)=cos(3x)f(x) = \text{cos}(3x)f(x)=cos(3x).

Answer: f′(x)=−3sin(3x)f'(x) = -3 \text{sin}(3x)f′(x)=−3sin(3x). Apply Chain Rule: derivative of cos⁡(u)\cos(u)cos(u) is −sin⁡(u)⋅u′-\sin(u) \cdot u'−sin(u)⋅u′ where u=3xu = 3xu=3x.

Flashcard 3: Which rule is used for differentiating compositions of functions?

Answer: The Chain Rule. Use this rule for composite functions like f(g(x))f(g(x))f(g(x)).

Flashcard 4: Find the derivative of 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 the Power Rule.

Flashcard 5: Find the derivative of f(x)=tan(x2)f(x) = \text{tan}(x^2)f(x)=tan(x2).

Answer: f′(x)=2xsec2(x2)f'(x) = 2x \text{sec}^2(x^2)f′(x)=2xsec2(x2). Apply Chain Rule: derivative of tan⁡(u)\tan(u)tan(u) is sec⁡2(u)⋅u′\sec^2(u) \cdot u'sec2(u)⋅u′ where u=x2u = x^2u=x2.

Flashcard 6: What is the geometric interpretation of a derivative?

Answer: The slope of the tangent line to the curve at a point. The derivative gives the instantaneous slope at any point.

Flashcard 7: Which function's derivative is exe^xex?

Answer: exe^xex. The exponential function exe^xex is its own derivative.

Flashcard 8: Find the derivative of f(x)=x4−2x2+xf(x) = x^4 - 2x^2 + xf(x)=x4−2x2+x.

Answer: f′(x)=4x3−4x+1f'(x) = 4x^3 - 4x + 1f′(x)=4x3−4x+1. Apply Power Rule to each term: 4x3−4x+14x^3 - 4x + 14x3−4x+1.

Flashcard 9: Identify the derivative of a constant function.

Answer: Zero. Constants have no rate of change, so their derivative is zero.

Flashcard 10: Find the derivative of 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 the reciprocal function.

Flashcard 11: Which rule is used to find the derivative of a product of two functions?

Answer: The Product Rule. Use this when differentiating products of two functions.

Flashcard 12: What is the derivative of arccos(x)\text{arccos}(x)arccos(x)?

Answer: −11−x2-\frac{1}{\sqrt{1-x^2}}−1−x2​1​. This is the standard derivative formula for inverse cosine.

Flashcard 13: What is the derivative of ln(f(x))\text{ln}(f(x))ln(f(x))?

Answer: f′(x)f(x)\frac{f'(x)}{f(x)}f(x)f′(x)​. Use the Chain Rule with the natural logarithm function.

Flashcard 14: What is the derivative of arcsin(x)\text{arcsin}(x)arcsin(x)?

Answer: 11−x2\frac{1}{\sqrt{1-x^2}}1−x2​1​. This is the standard derivative formula for inverse sine.

Flashcard 15: Find the derivative of f(x)=ex2f(x) = \text{e}^{x^2}f(x)=ex2.

Answer: f′(x)=2xex2f'(x) = 2x \text{e}^{x^2}f′(x)=2xex2. Apply Chain Rule: derivative of eue^ueu is eu⋅u′e^u \cdot u'eu⋅u′ where u=x2u = x^2u=x2.

Flashcard 16: What is the derivative of cos⁡(x)\cos(x)cos(x)?

Answer: -sin⁡(x)\sin(x)sin(x). The derivative of cosine is negative sine.

Flashcard 17: What is the derivative of sec(x)\text{sec}(x)sec(x)?

Answer: sec(x)tan(x)\text{sec}(x)\text{tan}(x)sec(x)tan(x). This is the standard derivative formula for secant.

Flashcard 18: Find the derivative of f(x)=xexf(x) = x \text{e}^xf(x)=xex using the Product Rule.

Answer: f′(x)=ex+xexf'(x) = \text{e}^x + x \text{e}^xf′(x)=ex+xex. Apply Product Rule: 1⋅ex+x⋅ex=ex(1+x)1 \cdot e^x + x \cdot e^x = e^x(1 + x)1⋅ex+x⋅ex=ex(1+x).

Flashcard 19: What is the derivative of cot(x)\text{cot}(x)cot(x)?

Answer: −csc2(x)-\text{csc}^2(x)−csc2(x). This is the standard derivative formula for cotangent.

Flashcard 20: Find the derivative of f(x)=ln(x2+1)f(x) = \text{ln}(x^2 + 1)f(x)=ln(x2+1).

Answer: f′(x)=2xx2+1f'(x) = \frac{2x}{x^2 + 1}f′(x)=x2+12x​. Apply Chain Rule: 1x2+1⋅2x\frac{1}{x^2 + 1} \cdot 2xx2+11​⋅2x.

Flashcard 21: What is the derivative of csc(x)\text{csc}(x)csc(x)?

Answer: −csc(x)cot(x)-\text{csc}(x)\text{cot}(x)−csc(x)cot(x). This is the standard derivative formula for cosecant.

Flashcard 22: Identify the derivative of f(x)=1sin(x)f(x) = \frac{1}{\text{sin}(x)}f(x)=sin(x)1​.

Answer: f′(x)=−cos(x)sin2(x)f'(x) = -\frac{\text{cos}(x)}{\text{sin}^2(x)}f′(x)=−sin2(x)cos(x)​. Rewrite as csc⁡(x)\csc(x)csc(x) and use the derivative formula.

Flashcard 23: State the derivative of arctan(x)\text{arctan}(x)arctan(x).

Answer: 11+x2\frac{1}{1+x^2}1+x21​. This is the standard derivative formula for inverse tangent.

Flashcard 24: What is the derivative of ef(x)\text{e}^{f(x)}ef(x)?

Answer: f′(x)ef(x)f'(x) \text{e}^{f(x)}f′(x)ef(x). Use the Chain Rule with the exponential function.

Flashcard 25: Identify the derivative of the inverse function f−1(x)f^{-1}(x)f−1(x).

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​. The derivative of an inverse function uses this reciprocal relationship.

Flashcard 26: Which rule is used to find the derivative of a quotient of two functions?

Answer: The Quotient Rule. Use this when differentiating quotients of two functions.

Flashcard 27: State the Product Rule for derivatives.

Answer: If u(x)u(x)u(x) and v(x)v(x)v(x), then (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′. First function times derivative of second plus second times derivative of first.

Flashcard 28: State the Power Rule for derivatives.

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

Flashcard 29: Find the derivative of f(x)=3x2+2x+1f(x) = 3x^2 + 2x + 1f(x)=3x2+2x+1.

Answer: f′(x)=6x+2f'(x) = 6x + 2f′(x)=6x+2. Apply the Power Rule to each term separately.

Flashcard 30: What is the definition of a derivative?

Answer: The limit of the average rate of change as the interval approaches zero. This captures the instantaneous rate of change concept.