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AP Calculus BC Flashcards: Derivatives Of Trigonometry And Logarithmic Functions

Study Derivatives Of Trigonometry And Logarithmic Functions 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 Derivatives Of Trigonometry And Logarithmic Functions, 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: Derivatives Of Trigonometry And Logarithmic Functions

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QUESTION

Find the second derivative of y=cos xy = \text{cos } xy=cos x.

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ANSWER

d2ydx2=−cos x\frac{d^2y}{dx^2} = -\text{cos } xdx2d2y​=−cos x. Differentiate f′(x)=−sin xf'(x) = -\text{sin } xf′(x)=−sin x to get f′′(x)=−cos xf''(x) = -\text{cos } xf′′(x)=−cos x.

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Flashcard 1: Find the second derivative of y=cos xy = \text{cos } xy=cos x.

Answer: d2ydx2=−cos x\frac{d^2y}{dx^2} = -\text{cos } xdx2d2y​=−cos x. Differentiate f′(x)=−sin xf'(x) = -\text{sin } xf′(x)=−sin x to get f′′(x)=−cos xf''(x) = -\text{cos } xf′′(x)=−cos x.

Flashcard 2: Determine the derivative of f(x)=xln xf(x) = x\text{ln } xf(x)=xln x.

Answer: f′(x)=ln x+1f'(x) = \text{ln } x + 1f′(x)=ln x+1. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′.

Flashcard 3: Determine f′(x)f'(x)f′(x) if f(x)=cos x+sin xf(x) = \text{cos } x + \text{sin } xf(x)=cos x+sin x.

Answer: f′(x)=−sin x+cos xf'(x) = -\text{sin } x + \text{cos } xf′(x)=−sin x+cos x. Use sum rule: derivative of each term separately.

Flashcard 4: Evaluate f′(x)f'(x)f′(x) for f(x)=cos x−exf(x) = \text{cos } x - e^xf(x)=cos x−ex at x=0x = 0x=0.

Answer: f′(0)=−1f'(0) = -1f′(0)=−1. f′(x)=−sin x−exf'(x) = -\text{sin } x - e^xf′(x)=−sin x−ex, so f′(0)=0−1=−1f'(0) = 0 - 1 = -1f′(0)=0−1=−1.

Flashcard 5: What is the derivative of cos x\text{cos } xcos x?

Answer: −sin x-\text{sin } x−sin x. Basic derivative rule for cosine function.

Flashcard 6: Find the derivative: y=cos xy = \text{cos } xy=cos x.

Answer: dydx=−sin x\frac{dy}{dx} = -\text{sin } xdxdy​=−sin x. Apply the derivative rule for cos x\text{cos } xcos x.

Flashcard 7: Find the derivative of f(x)=sin⁡(2x)f(x) = \sin(2x)f(x)=sin(2x).

Answer: f′(x)=2cos⁡(2x)f'(x) = 2\cos(2x)f′(x)=2cos(2x). Use chain rule with inner function 2x2x2x.

Flashcard 8: Find the second derivative of y=ln xy = \text{ln } xy=ln x.

Answer: d2ydx2=−1x2\frac{d^2y}{dx^2} = -\frac{1}{x^2}dx2d2y​=−x21​. Differentiate f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1​ to get f′′(x)=−1x2f''(x) = -\frac{1}{x^2}f′′(x)=−x21​.

Flashcard 9: Determine the derivative of f(x)=xsin xf(x) = x\text{sin } xf(x)=xsin x.

Answer: f′(x)=sin x+xcos xf'(x) = \text{sin } x + x\text{cos } xf′(x)=sin x+xcos x. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′.

Flashcard 10: Differentiate f(x)=3cos x−2sin xf(x) = 3\text{cos } x - 2\text{sin } xf(x)=3cos x−2sin x.

Answer: f′(x)=−3sin x−2cos xf'(x) = -3\text{sin } x - 2\text{cos } xf′(x)=−3sin x−2cos x. Use sum/difference rule with constant multiples.

Flashcard 11: Find the slope of the tangent to y=ln xy = \text{ln } xy=ln x at x=4x = 4x=4.

Answer: Slope = 14\frac{1}{4}41​. The slope equals the derivative at the point.

Flashcard 12: Evaluate f′(x)f'(x)f′(x) for f(x)=ex−ln xf(x) = e^x - \text{ln } xf(x)=ex−ln x at x=1x = 1x=1.

Answer: f′(1)=e−1f'(1) = e - 1f′(1)=e−1. f′(x)=ex−1xf'(x) = e^x - \frac{1}{x}f′(x)=ex−x1​, so f′(1)=e−1f'(1) = e - 1f′(1)=e−1.

Flashcard 13: Find the second derivative of y=exy = e^xy=ex.

Answer: d2ydx2=ex\frac{d^2y}{dx^2} = e^xdx2d2y​=ex. Differentiate f′(x)=exf'(x) = e^xf′(x)=ex to get f′′(x)=exf''(x) = e^xf′′(x)=ex.

Flashcard 14: Find the derivative of f(x)=ln (5x)f(x) = \text{ln }(5x)f(x)=ln (5x).

Answer: f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1​. Constant multiple rule: ddx[ln(5x)]=15x⋅5=1x\frac{d}{dx}[\text{ln}(5x)] = \frac{1}{5x} \cdot 5 = \frac{1}{x}dxd​[ln(5x)]=5x1​⋅5=x1​.

Flashcard 15: Find the derivative of f(x)=e2xf(x) = e^{2x}f(x)=e2x.

Answer: f′(x)=2e2xf'(x) = 2e^{2x}f′(x)=2e2x. Use chain rule with inner function 2x2x2x.

Flashcard 16: Find 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). Use chain rule with inner function 3x3x3x.

Flashcard 17: Determine the derivative of f(x)=xexf(x) = xe^xf(x)=xex.

Answer: f′(x)=ex+xexf'(x) = e^x + xe^xf′(x)=ex+xex. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′.

Flashcard 18: Determine the derivative of f(x)=xcos xf(x) = x\text{cos } xf(x)=xcos x.

Answer: f′(x)=cos x−xsin xf'(x) = \text{cos } x - x\text{sin } xf′(x)=cos x−xsin x. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′.

Flashcard 19: Find the derivative: y=ln xy = \text{ln } xy=ln x.

Answer: dydx=1x\frac{dy}{dx} = \frac{1}{x}dxdy​=x1​. Apply the derivative rule for ln x\text{ln } xln x.

Flashcard 20: Differentiate f(x)=5ex+7ln xf(x) = 5e^x + 7\text{ln } xf(x)=5ex+7ln x.

Answer: f′(x)=5ex+7xf'(x) = 5e^x + \frac{7}{x}f′(x)=5ex+x7​. Use sum rule with constant multiples.

Flashcard 21: Evaluate the derivative: f(x)=exf(x) = e^xf(x)=ex at x=ln 2x = \text{ln } 2x=ln 2.

Answer: f′(ln 2)=2f'(\text{ln } 2) = 2f′(ln 2)=2. f′(x)=exf'(x) = e^xf′(x)=ex, so f′(ln 2)=eln 2=2f'(\text{ln } 2) = e^{\text{ln } 2} = 2f′(ln 2)=eln 2=2.

Flashcard 22: Determine f′(x)f'(x)f′(x) if f(x)=ex+ln xf(x) = e^x + \text{ln } xf(x)=ex+ln x.

Answer: f′(x)=ex+1xf'(x) = e^x + \frac{1}{x}f′(x)=ex+x1​. Use sum rule: derivative of each term separately.

Flashcard 23: Evaluate the derivative: f(x)=cos xf(x) = \text{cos } xf(x)=cos x at x=π2x = \frac{\text{π}}{2}x=2π​.

Answer: f′(π2)=−1f'\big(\frac{\text{π}}{2}\big) = -1f′(2π​)=−1. f′(x)=−sin xf'(x) = -\text{sin } xf′(x)=−sin x, so f′(π2)=−sin(π2)=−1f'\big(\frac{\text{π}}{2}\big) = -\text{sin}\big(\frac{\text{π}}{2}\big) = -1f′(2π​)=−sin(2π​)=−1.

Flashcard 24: Evaluate the derivative: f(x)=ln xf(x) = \text{ln } xf(x)=ln x at x=1x = 1x=1.

Answer: f′(1)=1f'(1) = 1f′(1)=1. f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1​, so f′(1)=11=1f'(1) = \frac{1}{1} = 1f′(1)=11​=1.

Flashcard 25: Evaluate the derivative: f(x)=ln xf(x) = \text{ln } xf(x)=ln x at x=2x = 2x=2.

Answer: f′(2)=12f'(2) = \frac{1}{2}f′(2)=21​. f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1​, so f′(2)=12f'(2) = \frac{1}{2}f′(2)=21​.

Flashcard 26: Differentiate f(x)=exf(x) = e^xf(x)=ex at x=1x = 1x=1.

Answer: f′(1)=ef'(1) = ef′(1)=e. f′(x)=exf'(x) = e^xf′(x)=ex, so f′(1)=e1=ef'(1) = e^1 = ef′(1)=e1=e.

Flashcard 27: Differentiate f(x)=cos xf(x) = \text{cos } xf(x)=cos x at x=0x = 0x=0.

Answer: f′(0)=0f'(0) = 0f′(0)=0. f′(x)=−sin xf'(x) = -\text{sin } xf′(x)=−sin x, so f′(0)=−sin (0)=0f'(0) = -\text{sin }(0) = 0f′(0)=−sin (0)=0.

Flashcard 28: Find the derivative: y=exy = e^xy=ex.

Answer: dydx=ex\frac{dy}{dx} = e^xdxdy​=ex. Apply the derivative rule for exe^xex.

Flashcard 29: What is the derivative of ln x\text{ln } xln x?

Answer: 1x\frac{1}{x}x1​. Standard derivative of natural logarithm.

Flashcard 30: What is the derivative of exe^xex?

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