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AP Calculus BC Flashcards: Derivative Notation

Study Derivative Notation 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 Derivative Notation, 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: Derivative Notation

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QUESTION

What is the derivative of f(x)=ln⁡xf(x) = \ln xf(x)=lnx?

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ANSWER

1x\frac{1}{x}x1​. Natural log derivative is reciprocal function.

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All flashcards

Flashcard 1: What is the derivative of f(x)=ln⁡xf(x) = \ln xf(x)=lnx?

Answer: 1x\frac{1}{x}x1​. Natural log derivative is reciprocal function.

Flashcard 2: What is the derivative of f(x)=ln⁡xf(x) = \ln xf(x)=lnx?

Answer: 1x\frac{1}{x}x1​. Natural log derivative is reciprocal function.

Flashcard 3: What is the derivative of f(x)=cos⁡xf(x) = \cos xf(x)=cosx?

Answer: −sin⁡x-\sin x−sinx. Derivative of cosine is negative sine.

Flashcard 4: Find the derivative of f(x)=3x2+2xf(x) = 3x^2 + 2xf(x)=3x2+2x.

Answer: f′(x)=6x+2f'(x) = 6x + 2f′(x)=6x+2. Apply power rule to each term separately.

Flashcard 5: What is the derivative of f(x)=cot⁡xf(x) = \cot xf(x)=cotx?

Answer: −csc⁡2x- \csc^2 x−csc2x. Derivative of cotangent is negative cosecant squared.

Flashcard 6: What does the derivative represent geometrically?

Answer: Slope of the tangent line. Derivative gives instantaneous rate of change.

Flashcard 7: Find the derivative of f(x)=x3f(x) = \sqrt{x^3}f(x)=x3​.

Answer: f′(x)=32x1/2f'(x) = \frac{3}{2}x^{1/2}f′(x)=23​x1/2. Rewrite x3=x3/2\sqrt{x^3} = x^{3/2}x3​=x3/2 and use power rule.

Flashcard 8: Find the derivative of f(x)=12x−2f(x) = \frac{1}{2}x^{-2}f(x)=21​x−2.

Answer: f′(x)=−x−3f'(x) = -x^{-3}f′(x)=−x−3. Constant 12\frac{1}{2}21​ times power rule on x−2x^{-2}x−2.

Flashcard 9: Find the derivative of f(x)=xcos⁡xf(x) = x \cos xf(x)=xcosx.

Answer: f′(x)=cos⁡x−xsin⁡xf'(x) = \cos x - x \sin xf′(x)=cosx−xsinx. Product rule with u=xu = xu=x and v=cos⁡xv = \cos xv=cosx.

Flashcard 10: Find the derivative of f(x)=5ex−4f(x) = 5e^x - 4f(x)=5ex−4.

Answer: f′(x)=5exf'(x) = 5e^xf′(x)=5ex. Constant multiple rule with exponential derivative.

Flashcard 11: Find the derivative of f(x)=8x−1/2f(x) = 8x^{-1/2}f(x)=8x−1/2.

Answer: f′(x)=−4x−3/2f'(x) = -4x^{-3/2}f′(x)=−4x−3/2. Constant 8 times power rule on x−1/2x^{-1/2}x−1/2.

Flashcard 12: Find the derivative of f(x)=13x3f(x) = \frac{1}{3}x^3f(x)=31​x3.

Answer: f′(x)=x2f'(x) = x^2f′(x)=x2. Constant factor 13\frac{1}{3}31​ remains, apply power rule.

Flashcard 13: Find the derivative of f(x)=2sec⁡xf(x) = 2\sec xf(x)=2secx.

Answer: f′(x)=2sec⁡xtan⁡xf'(x) = 2\sec x \tan xf′(x)=2secxtanx. Constant multiple of secant derivative.

Flashcard 14: 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 use power rule.

Flashcard 15: What is the derivative of f(x)=csc⁡xf(x) = \csc xf(x)=cscx?

Answer: −csc⁡xcot⁡x - \csc x \cot x−cscxcotx. Derivative is negative product of cosecant and cotangent.

Flashcard 16: Find the derivative of f(x)=x2+1xf(x) = \frac{x^2 + 1}{x}f(x)=xx2+1​.

Answer: f′(x)=x2−1x2f'(x) = \frac{x^2 - 1}{x^2}f′(x)=x2x2−1​. Rewrite as x+x−1x + x^{-1}x+x−1 and differentiate.

Flashcard 17: What is the derivative notation using Leibniz's notation for y=f(x)y = f(x)y=f(x)?

Answer: dydx\frac{dy}{dx}dxdy​. Leibniz notation shows derivative of yyy with respect to xxx.

Flashcard 18: Find the derivative of f(x)=x4−3x2+xf(x) = x^4 - 3x^2 + xf(x)=x4−3x2+x.

Answer: f′(x)=4x3−6x+1f'(x) = 4x^3 - 6x + 1f′(x)=4x3−6x+1. Apply power rule to each term.

Flashcard 19: Find the derivative of f(x)=ln⁡(x2+1)f(x) = \ln(x^2 + 1)f(x)=ln(x2+1).

Answer: 2xx2+1\frac{2x}{x^2 + 1}x2+12x​. Use chain rule with ln⁡\lnln and x2+1x^2 + 1x2+1.

Flashcard 20: Find the derivative of f(x)=xf(x) = \sqrt{x}f(x)=x​.

Answer: f′(x)=12xf'(x) = \frac{1}{2\sqrt{x}}f′(x)=2x​1​. Rewrite as x1/2x^{1/2}x1/2 and apply power rule.

Flashcard 21: What is the derivative of f(x)=tan⁡xf(x) = \tan xf(x)=tanx?

Answer: sec⁡2x\sec^2 xsec2x. Derivative of tangent is secant squared.

Flashcard 22: What is the limit definition of the derivative of f(x)f(x)f(x) at x=ax = ax=a?

Answer: f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{{h \to 0}} \frac{f(a+h) - f(a)}{h}f′(a)=limh→0​hf(a+h)−f(a)​. Limit of difference quotient as hhh approaches 0.

Flashcard 23: What is the derivative of f(x)=axf(x) = a^xf(x)=ax where a>0a > 0a>0?

Answer: axln⁡aa^x \ln aaxlna. Exponential with base aaa requires ln⁡a\ln alna factor.

Flashcard 24: Find the derivative of f(x)=x3−5x+4f(x) = x^3 - 5x + 4f(x)=x3−5x+4.

Answer: f′(x)=3x2−5f'(x) = 3x^2 - 5f′(x)=3x2−5. Derivative of constant is 0, apply power rule to other terms.

Flashcard 25: Find the derivative of f(x)=ln⁡(sin⁡x)f(x) = \ln(\sin x)f(x)=ln(sinx).

Answer: f′(x)=cot⁡xf'(x) = \cot xf′(x)=cotx. Chain rule with ln⁡(sin⁡x)\ln(\sin x)ln(sinx).

Flashcard 26: Find the derivative of f(x)=tan⁡(x2)f(x) = \tan(x^2)f(x)=tan(x2).

Answer: f′(x)=2xsec⁡2(x2)f'(x) = 2x \sec^2(x^2)f′(x)=2xsec2(x2). Chain rule: outer derivative times inner derivative.

Flashcard 27: State the formula for the derivative of f(x)=xnf(x) = x^nf(x)=xn.

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

Flashcard 28: Find the derivative of f(x)=sin⁡2xf(x) = \sin^2 xf(x)=sin2x.

Answer: f′(x)=2sin⁡xcos⁡xf'(x) = 2\sin x \cos xf′(x)=2sinxcosx. Use chain rule with sin⁡2x=(sin⁡x)2\sin^2 x = (\sin x)^2sin2x=(sinx)2.

Flashcard 29: Find the derivative of f(x)=7x5f(x) = 7x^5f(x)=7x5.

Answer: f′(x)=35x4f'(x) = 35x^4f′(x)=35x4. Constant multiple of power rule.

Flashcard 30: Find the derivative of f(x)=x2sin⁡xf(x) = x^2 \sin xf(x)=x2sinx.

Answer: 2xsin⁡x+x2cos⁡x2x \sin x + x^2 \cos x2xsinx+x2cosx. Apply product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′.