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AP Calculus BC Flashcards: Selecting Procedures For Calculating Derivatives

Study Selecting Procedures For Calculating Derivatives 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 Selecting Procedures For Calculating Derivatives, 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: Selecting Procedures For Calculating Derivatives

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QUESTION

Find the derivative of f(x)=ln(sin(x))f(x) = \text{ln}(\text{sin}(x))f(x)=ln(sin(x)).

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ANSWER

f′(x)=cos(x)sin(x)f'(x) = \frac{\text{cos}(x)}{\text{sin}(x)}f′(x)=sin(x)cos(x)​. Chain rule: 1sin⁡(x)⋅cos⁡(x)\frac{1}{\sin(x)} \cdot \cos(x)sin(x)1​⋅cos(x).

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Flashcard 1: Find the derivative of f(x)=ln(sin(x))f(x) = \text{ln}(\text{sin}(x))f(x)=ln(sin(x)).

Answer: f′(x)=cos(x)sin(x)f'(x) = \frac{\text{cos}(x)}{\text{sin}(x)}f′(x)=sin(x)cos(x)​. Chain rule: 1sin⁡(x)⋅cos⁡(x)\frac{1}{\sin(x)} \cdot \cos(x)sin(x)1​⋅cos(x).

Flashcard 2: Identify the product rule for derivatives.

Answer: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′. Product rule: first times derivative of second plus second times derivative of first.

Flashcard 3: State the quotient rule for derivatives.

Answer: (u/v)′=u′v−uv′v2(u/v)' = \frac{u'v - uv'}{v^2}(u/v)′=v2u′v−uv′​. Quotient rule formula for division of functions.

Flashcard 4: Determine the derivative of f(x)=sin2(x)f(x) = \text{sin}^2(x)f(x)=sin2(x).

Answer: f′(x)=2sin(x)cos(x)f'(x) = 2\text{sin}(x)\text{cos}(x)f′(x)=2sin(x)cos(x). Chain rule: 2sin⁡(x)cos⁡(x)2\sin(x)\cos(x)2sin(x)cos(x).

Flashcard 5: What is the derivative of cos(x)\text{cos}(x)cos(x) with respect to xxx?

Answer: ddx[cos(x)]=−sin(x)\frac{d}{dx}[\text{cos}(x)] = -\text{sin}(x)dxd​[cos(x)]=−sin(x). Derivative of cosine is negative sine.

Flashcard 6: What is 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). Chain rule with tangent: sec⁡2(x2)⋅2x\sec^2(x^2) \cdot 2xsec2(x2)⋅2x.

Flashcard 7: State the derivative of tan(x)\text{tan}(x)tan(x) with respect to xxx.

Answer: ddx[tan(x)]=sec2(x)\frac{d}{dx}[\text{tan}(x)] = \text{sec}^2(x)dxd​[tan(x)]=sec2(x). Derivative of tangent is secant squared.

Flashcard 8: 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 exponent, reduce power by 1.

Flashcard 9: What is the derivative of csc(x)\text{csc}(x)csc(x) with respect to xxx?

Answer: ddx[csc(x)]=−csc(x)cot(x)\frac{d}{dx}[\text{csc}(x)] = -\text{csc}(x)\text{cot}(x)dxd​[csc(x)]=−csc(x)cot(x). Cosecant derivative involves cotangent.

Flashcard 10: What is the derivative of cot(x)\text{cot}(x)cot(x) with respect to xxx?

Answer: ddx[cot(x)]=−csc2(x)\frac{d}{dx}[\text{cot}(x)] = -\text{csc}^2(x)dxd​[cot(x)]=−csc2(x). Cotangent derivative is negative cosecant squared.

Flashcard 11: Find the derivative of f(x)=sec2(x)f(x) = \text{sec}^2(x)f(x)=sec2(x).

Answer: f′(x)=2sec2(x)sec(x)tan(x)f'(x) = 2\text{sec}^2(x)\text{sec}(x)\text{tan}(x)f′(x)=2sec2(x)sec(x)tan(x). Chain rule: 2sec⁡(x)⋅sec⁡(x)tan⁡(x)2\sec(x) \cdot \sec(x)\tan(x)2sec(x)⋅sec(x)tan(x).

Flashcard 12: What is 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​. Chain rule: 1x2+1⋅2x\frac{1}{x^2+1} \cdot 2xx2+11​⋅2x.

Flashcard 13: Calculate the derivative of f(x)=cos3(x)f(x) = \text{cos}^3(x)f(x)=cos3(x).

Answer: f′(x)=−3cos2(x)sin(x)f'(x) = -3\text{cos}^2(x)\text{sin}(x)f′(x)=−3cos2(x)sin(x). Chain rule: 3cos⁡2(x)⋅(−sin⁡(x))3\cos^2(x) \cdot (-\sin(x))3cos2(x)⋅(−sin(x)).

Flashcard 14: What is the derivative of f(x)=esin(x)f(x) = \text{e}^{\text{sin}(x)}f(x)=esin(x)?

Answer: f′(x)=esin(x)cos(x)f'(x) = \text{e}^{\text{sin}(x)}\text{cos}(x)f′(x)=esin(x)cos(x). Chain rule: esin⁡(x)⋅cos⁡(x)e^{\sin(x)} \cdot \cos(x)esin(x)⋅cos(x).

Flashcard 15: Determine the derivative of f(x)=exxf(x) = \frac{\text{e}^x}{x}f(x)=xex​.

Answer: f′(x)=ex(x−1)x2f'(x) = \frac{\text{e}^x(x-1)}{x^2}f′(x)=x2ex(x−1)​. Quotient rule with u=ex,v=xu = e^x, v = xu=ex,v=x.

Flashcard 16: Calculate the derivative of f(x)=sin(x)+cos(x)f(x) = \text{sin}(x) + \text{cos}(x)f(x)=sin(x)+cos(x).

Answer: f′(x)=cos(x)−sin(x)f'(x) = \text{cos}(x) - \text{sin}(x)f′(x)=cos(x)−sin(x). Sum rule: derivative of sum equals sum of derivatives.

Flashcard 17: Identify the derivative of f(x)=csc(x3)f(x) = \text{csc}(x^3)f(x)=csc(x3).

Answer: f′(x)=−3x2csc(x3)cot(x3)f'(x) = -3x^2\text{csc}(x^3)\text{cot}(x^3)f′(x)=−3x2csc(x3)cot(x3). Chain rule with cosecant function.

Flashcard 18: Compute the derivative of f(x)=1x2+1f(x) = \frac{1}{x^2 + 1}f(x)=x2+11​.

Answer: f′(x)=−2x(x2+1)2f'(x) = -\frac{2x}{(x^2 + 1)^2}f′(x)=−(x2+1)22x​. Quotient rule with u=1,v=x2+1u = 1, v = x^2 + 1u=1,v=x2+1.

Flashcard 19: 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. Chain rule: ex2⋅2xe^{x^2} \cdot 2xex2⋅2x.

Flashcard 20: Find the derivative of f(x)=x2sin(x)f(x) = x^2 \text{sin}(x)f(x)=x2sin(x).

Answer: f′(x)=2xsin(x)+x2cos(x)f'(x) = 2x\text{sin}(x) + x^2\text{cos}(x)f′(x)=2xsin(x)+x2cos(x). Product rule: u′v+uv′u'v + uv'u′v+uv′ where u=x2,v=sin⁡(x)u = x^2, v = \sin(x)u=x2,v=sin(x).

Flashcard 21: Find the derivative of f(x)=e2xf(x) = e^{2x}f(x)=e2x with respect to xxx.

Answer: f′(x)=2e2xf'(x) = 2e^{2x}f′(x)=2e2x. Chain rule: derivative of inside times e2xe^{2x}e2x.

Flashcard 22: Find the derivative of f(x)=3x4+2x2−5f(x) = 3x^4 + 2x^2 - 5f(x)=3x4+2x2−5.

Answer: f′(x)=12x3+4xf'(x) = 12x^3 + 4xf′(x)=12x3+4x. Apply power rule to each term separately.

Flashcard 23: What is the derivative of sec(x)\text{sec}(x)sec(x) with respect to xxx?

Answer: ddx[sec(x)]=sec(x)tan(x)\frac{d}{dx}[\text{sec}(x)] = \text{sec}(x)\text{tan}(x)dxd​[sec(x)]=sec(x)tan(x). Secant derivative involves tangent.

Flashcard 24: Determine the derivative of f(x)=x4+3x−2f(x) = x^4 + 3x^{-2}f(x)=x4+3x−2.

Answer: f′(x)=4x3−6x−3f'(x) = 4x^3 - 6x^{-3}f′(x)=4x3−6x−3. Power rule applied to positive and negative exponents.

Flashcard 25: What is the derivative of f(x)=sec(2x)f(x) = \text{sec}(2x)f(x)=sec(2x)?

Answer: f′(x)=2sec(2x)tan(2x)f'(x) = 2\text{sec}(2x)\text{tan}(2x)f′(x)=2sec(2x)tan(2x). Chain rule with secant function.

Flashcard 26: What is the derivative of f(x)=ln(2x)f(x) = \text{ln}(2x)f(x)=ln(2x)?

Answer: f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1​. Chain rule: ddx[ln⁡(2x)]=12x⋅2=1x\frac{d}{dx}[\ln(2x)] = \frac{1}{2x} \cdot 2 = \frac{1}{x}dxd​[ln(2x)]=2x1​⋅2=x1​.

Flashcard 27: Identify the derivative of f(x)=x5/3f(x) = x^{5/3}f(x)=x5/3.

Answer: f′(x)=53x2/3f'(x) = \frac{5}{3}x^{2/3}f′(x)=35​x2/3. Power rule with fractional exponent.

Flashcard 28: State the formula for the derivative of ln(x)\text{ln}(x)ln(x).

Answer: ddx[ln(x)]=1x\frac{d}{dx}[\text{ln}(x)] = \frac{1}{x}dxd​[ln(x)]=x1​. Natural log derivative is reciprocal function.

Flashcard 29: State the formula for the derivative of sin(x)\text{sin}(x)sin(x).

Answer: ddx[sin(x)]=cos(x)\frac{d}{dx}[\text{sin}(x)] = \text{cos}(x)dxd​[sin(x)]=cos(x). Derivative of sine is cosine.

Flashcard 30: What is the derivative of f(x)=xnf(x) = x^nf(x)=xn with respect to xxx?

Answer: f′(x)=nxn−1f'(x) = nx^{n-1}f′(x)=nxn−1. Power rule: bring down exponent, reduce power by 1.