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.

AP Calculus BC

Derivatives Of Trigonometry And Logarithmic Functions

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QUESTION
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Evaluate f(x)f'(x) for f(x)=cos xexf(x) = \text{cos } x - e^x at x=0x = 0.

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ANSWER

f(0)=1f'(0) = -1. f(x)=sin xexf'(x) = -\text{sin } x - e^x, so f(0)=01=1f'(0) = 0 - 1 = -1.

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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.

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Flashcard 1: Evaluate f(x)f'(x) for f(x)=cos xexf(x) = \text{cos } x - e^x at x=0x = 0.

Answer: f(0)=1f'(0) = -1. f(x)=sin xexf'(x) = -\text{sin } x - e^x, so f(0)=01=1f'(0) = 0 - 1 = -1.

Flashcard 2: Find the second derivative of y=sin xy = \text{sin } x.

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

Flashcard 3: Find the derivative: y=ln xy = \text{ln } x.

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

Flashcard 4: Find the derivative of f(x)=e2xf(x) = e^{2x}.

Answer: f(x)=2e2xf'(x) = 2e^{2x}. Use chain rule with inner function 2x2x.

Flashcard 5: Evaluate the derivative: f(x)=ln xf(x) = \text{ln } x at x=2x = 2.

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

Flashcard 6: Find the second derivative of y=ln xy = \text{ln } x.

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

Flashcard 7: Evaluate the derivative: f(x)=ln xf(x) = \text{ln } x at x=1x = 1.

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

Flashcard 8: Differentiate f(x)=exf(x) = e^x at x=1x = 1.

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

Flashcard 9: Find the derivative of f(x)=ln (5x)f(x) = \text{ln }(5x).

Answer: f(x)=1xf'(x) = \frac{1}{x}. Constant multiple rule: ddx[ln(5x)]=15x5=1x\frac{d}{dx}[\text{ln}(5x)] = \frac{1}{5x} \cdot 5 = \frac{1}{x}.

Flashcard 10: Evaluate the derivative: f(x)=sin xf(x) = \text{sin } x at x=0x = 0.

Answer: f(0)=1f'(0) = 1. f(x)=cos xf'(x) = \text{cos } x, so f(0)=cos(0)=1f'(0) = \text{cos}(0) = 1.

Flashcard 11: Find the second derivative of y=exy = e^x.

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

Flashcard 12: Determine the derivative of f(x)=xexf(x) = xe^x.

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

Flashcard 13: Determine f(x)f'(x) if f(x)=cos x+sin xf(x) = \text{cos } x + \text{sin } x.

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

Flashcard 14: Differentiate f(x)=sin xf(x) = \text{sin } x at x=π2x = \frac{\text{π}}{2}.

Answer: f(π2)=0f'\big(\frac{\text{π}}{2}\big) = 0. f(x)=cos xf'(x) = \text{cos } x, so f(π2)=cos(π2)=0f'\big(\frac{\text{π}}{2}\big) = \text{cos}\big(\frac{\text{π}}{2}\big) = 0.

Flashcard 15: Differentiate f(x)=sin xf(x) = \text{sin } x at x=π2x = \frac{\text{π}}{2}.

Answer: f(π2)=0f'\big(\frac{\text{π}}{2}\big) = 0. f(x)=cos xf'(x) = \text{cos } x, so f(π2)=cos(π2)=0f'\big(\frac{\text{π}}{2}\big) = \text{cos}\big(\frac{\text{π}}{2}\big) = 0.

Flashcard 16: Evaluate the derivative: f(x)=cos xf(x) = \text{cos } x at x=π2x = \frac{\text{π}}{2}.

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

Flashcard 17: What is the derivative of sin x\text{sin } x?

Answer: cos x\text{cos } x. Basic derivative rule for sine function.

Flashcard 18: Find the derivative of f(x)=e2xf(x) = e^{2x}.

Answer: f(x)=2e2xf'(x) = 2e^{2x}. Use chain rule with inner function 2x2x.

Flashcard 19: Find the derivative of f(x)=cos (3x)f(x) = \text{cos }(3x).

Answer: f(x)=3sin (3x)f'(x) = -3\text{sin }(3x). Use chain rule with inner function 3x3x.

Flashcard 20: Find the derivative: y=cos xy = \text{cos } x.

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

Flashcard 21: Find the slope of the tangent to y=ln xy = \text{ln } x at x=4x = 4.

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

Flashcard 22: Find the derivative: y=ln xy = \text{ln } x.

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

Flashcard 23: Find the derivative of f(x)=sin (2x)f(x) = \text{sin }(2x).

Answer: f(x)=2cos (2x)f'(x) = 2\text{cos }(2x). Use chain rule with inner function 2x2x.

Flashcard 24: Find the second derivative of y=cos xy = \text{cos } x.

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

Flashcard 25: Differentiate f(x)=ln xf(x) = \text{ln } x at x=ex = e.

Answer: f(e)=1ef'(e) = \frac{1}{e}. f(x)=1xf'(x) = \frac{1}{x}, so f(e)=1ef'(e) = \frac{1}{e}.

Flashcard 26: Find the derivative of f(x)=ln (5x)f(x) = \text{ln }(5x).

Answer: f(x)=1xf'(x) = \frac{1}{x}. Constant multiple rule: ddx[ln(5x)]=15x5=1x\frac{d}{dx}[\text{ln}(5x)] = \frac{1}{5x} \cdot 5 = \frac{1}{x}.

Flashcard 27: What is the derivative of exe^x?

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

Flashcard 28: Differentiate f(x)=5ex+7ln xf(x) = 5e^x + 7\text{ln } x.

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

Flashcard 29: Determine the derivative of f(x)=xcos xf(x) = x\text{cos } x.

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

Flashcard 30: Find the derivative: y=exy = e^x.

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

Flashcard 31: Evaluate f(x)f'(x) for f(x)=exln xf(x) = e^x - \text{ln } x at x=1x = 1.

Answer: f(1)=e1f'(1) = e - 1. f(x)=ex1xf'(x) = e^x - \frac{1}{x}, so f(1)=e1f'(1) = e - 1.

Flashcard 32: Determine f(x)f'(x) if f(x)=cosx+sinxf(x) = \cos x + \sin x.

Answer: f(x)=sinx+cosxf'(x) = -\sin x + \cos x. Use sum rule: derivative of each term separately.

Flashcard 33: Evaluate the derivative: f(x)=ln xf(x) = \text{ln } x at x=2x = 2.

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

Flashcard 34: Determine f(x)f'(x) if f(x)=ex+ln xf(x) = e^x + \text{ln } x.

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

Flashcard 35: Differentiate f(x)=cos xf(x) = \text{cos } x at x=0x = 0.

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

Flashcard 36: Find the slope of the tangent to y=exy = e^x at x=2x = 2.

Answer: Slope = e2e^2. The slope equals the derivative at the point.

Flashcard 37: Evaluate the derivative: f(x)=exf(x) = e^x at x=0x = 0.

Answer: f(0)=1f'(0) = 1. f(x)=exf'(x) = e^x, so f(0)=e0=1f'(0) = e^0 = 1.

Flashcard 38: What is the derivative of ln x\text{ln } x?

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

Flashcard 39: Evaluate the derivative: f(x)=exf(x) = e^x at x=0x = 0.

Answer: f(0)=1f'(0) = 1. f(x)=exf'(x) = e^x, so f(0)=e0=1f'(0) = e^0 = 1.

Flashcard 40: What is the derivative of ln x\text{ln } x?

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

Flashcard 41: Determine the derivative of f(x)=xln xf(x) = x\text{ln } x.

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

Flashcard 42: Determine the derivative of f(x)=xcos xf(x) = x\text{cos } x.

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

Flashcard 43: Determine the derivative of f(x)=xsin xf(x) = x\text{sin } x.

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

Flashcard 44: Differentiate f(x)=3cos x2sin xf(x) = 3\text{cos } x - 2\text{sin } x.

Answer: f(x)=3sin x2cos xf'(x) = -3\text{sin } x - 2\text{cos } x. Use sum/difference rule with constant multiples.

Flashcard 45: Evaluate f(x)f'(x) for f(x)=exln xf(x) = e^x - \text{ln } x at x=1x = 1.

Answer: f(1)=e1f'(1) = e - 1. f(x)=ex1xf'(x) = e^x - \frac{1}{x}, so f(1)=e1f'(1) = e - 1.

Flashcard 46: Find the derivative: y=exy = e^x.

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

Flashcard 47: Evaluate the derivative: f(x)=exf(x) = e^x at x=ln 2x = \text{ln } 2.

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

Flashcard 48: Differentiate f(x)=3cos x2sin xf(x) = 3\text{cos } x - 2\text{sin } x.

Answer: f(x)=3sin x2cos xf'(x) = -3\text{sin } x - 2\text{cos } x. Use sum/difference rule with constant multiples.

Flashcard 49: Find the slope of the tangent to y=ln xy = \text{ln } x at x=4x = 4.

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

Flashcard 50: Find the second derivative of y=exy = e^x.

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

Flashcard 51: What is the derivative of cos x\text{cos } x?

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

Flashcard 52: Differentiate f(x)=cos xf(x) = \text{cos } x at x=0x = 0.

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

Flashcard 53: Differentiate f(x)=5ex+7ln xf(x) = 5e^x + 7\text{ln } x.

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

Flashcard 54: Find the derivative of f(x)=sin(2x)f(x) = \sin(2x).

Answer: f(x)=2cos(2x)f'(x) = 2\cos(2x). Use chain rule with inner function 2x2x.

Flashcard 55: Evaluate the derivative: f(x)=sin xf(x) = \text{sin } x at x=0x = 0.

Answer: f(0)=1f'(0) = 1. f(x)=cos xf'(x) = \text{cos } x, so f(0)=cos(0)=1f'(0) = \text{cos}(0) = 1.

Flashcard 56: What is the derivative of exe^x?

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

Flashcard 57: Determine the derivative of f(x)=xln xf(x) = x\text{ln } x.

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

Flashcard 58: Evaluate f(x)f'(x) for f(x)=cos xexf(x) = \text{cos } x - e^x at x=0x = 0.

Answer: f(0)=1f'(0) = -1. f(x)=sin xexf'(x) = -\text{sin } x - e^x, so f(0)=01=1f'(0) = 0 - 1 = -1.

Flashcard 59: Determine the derivative of f(x)=xexf(x) = xe^x.

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

Flashcard 60: Find the second derivative of y=cos xy = \text{cos } x.

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

Flashcard 61: Find the second derivative of y=sin xy = \text{sin } x.

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

Flashcard 62: What is the derivative of cos x\text{cos } x?

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

Flashcard 63: Evaluate the derivative: f(x)=ln xf(x) = \text{ln } x at x=1x = 1.

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

Flashcard 64: Find the second derivative of y=ln xy = \text{ln } x.

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

Flashcard 65: Find the derivative: y=sin xy = \text{sin } x.

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

Flashcard 66: Determine f(x)f'(x) if f(x)=ex+ln xf(x) = e^x + \text{ln } x.

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

Flashcard 67: What is the derivative of sin x\text{sin } x?

Answer: cos x\text{cos } x. Basic derivative rule for sine function.

Flashcard 68: Find the derivative: y=sin xy = \text{sin } x.

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

Flashcard 69: Differentiate f(x)=exf(x) = e^x at x=1x = 1.

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

Flashcard 70: Differentiate f(x)=ln xf(x) = \text{ln } x at x=ex = e.

Answer: f(e)=1ef'(e) = \frac{1}{e}. f(x)=1xf'(x) = \frac{1}{x}, so f(e)=1ef'(e) = \frac{1}{e}.

Flashcard 71: Find the derivative of f(x)=cos (3x)f(x) = \text{cos }(3x).

Answer: f(x)=3sin (3x)f'(x) = -3\text{sin }(3x). Use chain rule with inner function 3x3x.

Flashcard 72: Find the slope of the tangent to y=exy = e^x at x=2x = 2.

Answer: Slope = e2e^2. The slope equals the derivative at the point.

Flashcard 73: Evaluate the derivative: f(x)=cos xf(x) = \text{cos } x at x=π2x = \frac{\text{π}}{2}.

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

Flashcard 74: Find the derivative: y=cos xy = \text{cos } x.

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

Flashcard 75: Determine the derivative of f(x)=xsin xf(x) = x\text{sin } x.

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

Flashcard 76: Evaluate the derivative: f(x)=exf(x) = e^x at x=ln 2x = \text{ln } 2.

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