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  2. AP Calculus AB
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AP Calculus AB Flashcards: Derivatives Of Reciprocal Trig Functions

Study Derivatives Of Reciprocal Trig Functions in AP Calculus AB 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 Reciprocal Trig Functions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

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 AB Flashcards: Derivatives Of Reciprocal Trig Functions

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QUESTION

State the derivative of f(x)=cot⁡(3x3)f(x) = \cot(3x^3)f(x)=cot(3x3).

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ANSWER

−9x2csc⁡2(3x3)-9x^2\csc^2(3x^3)−9x2csc2(3x3). Chain rule: multiply by derivative of inner function 9x29x^29x2.

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Flashcard 1: State the derivative of f(x)=cot⁡(3x3)f(x) = \cot(3x^3)f(x)=cot(3x3).

Answer: −9x2csc⁡2(3x3)-9x^2\csc^2(3x^3)−9x2csc2(3x3). Chain rule: multiply by derivative of inner function 9x29x^29x2.

Flashcard 2: State the derivative of cot⁡(x)\cot(x)cot(x).

Answer: −csc⁡2(x)-\csc^2(x)−csc2(x). Standard derivative formula for cotangent function.

Flashcard 3: State the derivative of g(x)=sec⁡(x3)g(x) = \sec(x^3)g(x)=sec(x3).

Answer: 3x2sec⁡(x3)tan⁡(x3)3x^2\sec(x^3)\tan(x^3)3x2sec(x3)tan(x3). Chain rule: multiply by derivative of inner function 3x23x^23x2.

Flashcard 4: Compute d/dxd/dxd/dx for g(x)=sec⁡(x2+1)g(x) = \sec(x^2 + 1)g(x)=sec(x2+1).

Answer: 2xsec⁡(x2+1)tan⁡(x2+1)2x\sec(x^2 + 1)\tan(x^2 + 1)2xsec(x2+1)tan(x2+1). Chain rule: multiply by derivative of inner function 2x2x2x.

Flashcard 5: What is the derivative of f(x)=cot⁡(3x)f(x) = \cot(3x)f(x)=cot(3x)?

Answer: −3csc⁡2(3x)-3\csc^2(3x)−3csc2(3x). Chain rule: multiply by derivative of inner function 333.

Flashcard 6: What is the derivative of sec⁡(x)\sec(x)sec(x)?

Answer: sec⁡(x)tan⁡(x)\sec(x)\tan(x)sec(x)tan(x). Standard derivative formula for secant function.

Flashcard 7: State d/dxd/dxd/dx for g(x)=cot⁡(x)g(x) = \cot(\sqrt{x})g(x)=cot(x​).

Answer: −12xcsc⁡2(x)-\frac{1}{2\sqrt{x}}\csc^2(\sqrt{x})−2x​1​csc2(x​). Chain rule: multiply by derivative of x\sqrt{x}x​ which is 12x\frac{1}{2\sqrt{x}}2x​1​.

Flashcard 8: Find d/dxd/dxd/dx for g(x)=cot⁡(ln⁡(x))g(x) = \cot(\ln(x))g(x)=cot(ln(x)).

Answer: −1xcsc⁡2(ln⁡(x))-\frac{1}{x}\csc^2(\ln(x))−x1​csc2(ln(x)). Chain rule: multiply by derivative of ln⁡(x)\ln(x)ln(x) which is 1x\frac{1}{x}x1​.

Flashcard 9: Compute the derivative of g(x)=csc⁡(x4)g(x) = \csc(x^4)g(x)=csc(x4).

Answer: −4x3csc⁡(x4)cot⁡(x4)-4x^3\csc(x^4)\cot(x^4)−4x3csc(x4)cot(x4). Chain rule: multiply by derivative of inner function 4x34x^34x3.

Flashcard 10: Determine the derivative of y=tan⁡(3x2)y = \tan(3x^2)y=tan(3x2).

Answer: 6xsec⁡2(3x2)6x\sec^2(3x^2)6xsec2(3x2). Chain rule: multiply by derivative of inner function 6x6x6x.

Flashcard 11: What is the derivative of f(x)=tan⁡(x2+1)f(x) = \tan(x^2 + 1)f(x)=tan(x2+1)?

Answer: 2xsec⁡2(x2+1)2x\sec^2(x^2 + 1)2xsec2(x2+1). Chain rule: multiply by derivative of inner function 2x2x2x.

Flashcard 12: Identify the derivative of y=tan⁡(x+π)y = \tan(x + \pi)y=tan(x+π).

Answer: sec⁡2(x+π)\sec^2(x + \pi)sec2(x+π). Chain rule: multiply by derivative of inner function 111.

Flashcard 13: Determine d/dxd/dxd/dx for y=csc⁡(ex)y = \csc(e^x)y=csc(ex).

Answer: −excsc⁡(ex)cot⁡(ex)-e^x\csc(e^x)\cot(e^x)−excsc(ex)cot(ex). Chain rule: multiply by derivative of exe^xex which is exe^xex.

Flashcard 14: Find d/dxd/dxd/dx for h(x)=sec⁡(3x+1)h(x) = \sec(3x + 1)h(x)=sec(3x+1).

Answer: 3sec⁡(3x+1)tan⁡(3x+1)3\sec(3x + 1)\tan(3x + 1)3sec(3x+1)tan(3x+1). Chain rule: multiply by derivative of inner function 333.

Flashcard 15: Calculate the derivative of y=csc⁡(4x)y = \csc(4x)y=csc(4x).

Answer: −4csc⁡(4x)cot⁡(4x)-4\csc(4x)\cot(4x)−4csc(4x)cot(4x). Chain rule: multiply by derivative of inner function 444.

Flashcard 16: State the derivative of g(x)=cot⁡(2x3)g(x) = \cot(2x^3)g(x)=cot(2x3).

Answer: −6x2csc⁡2(2x3)-6x^2\csc^2(2x^3)−6x2csc2(2x3). Chain rule: multiply by derivative of inner function 6x26x^26x2.

Flashcard 17: What is the derivative of f(x)=tan⁡(5x2)f(x) = \tan(5x^2)f(x)=tan(5x2)?

Answer: 10xsec⁡2(5x2)10x\sec^2(5x^2)10xsec2(5x2). Chain rule: multiply by derivative of inner function 10x10x10x.

Flashcard 18: State the derivative of f(x)=cot⁡(3x3)f(x) = \cot(3x^3)f(x)=cot(3x3).

Answer: −9x2csc⁡2(3x3)-9x^2\csc^2(3x^3)−9x2csc2(3x3). Chain rule: multiply by derivative of inner function 9x29x^29x2.

Flashcard 19: Find the derivative of h(x)=csc⁡(ln⁡(x))h(x) = \csc(\ln(x))h(x)=csc(ln(x)).

Answer: −1xcsc⁡(ln⁡(x))cot⁡(ln⁡(x))-\frac{1}{x}\csc(\ln(x))\cot(\ln(x))−x1​csc(ln(x))cot(ln(x)). Chain rule: multiply by derivative of ln⁡(x)\ln(x)ln(x) which is 1x\frac{1}{x}x1​.

Flashcard 20: Calculate d/dxd/dxd/dx for g(x)=sec⁡(x5)g(x) = \sec(x^5)g(x)=sec(x5).

Answer: 5x4sec⁡(x5)tan⁡(x5)5x^4\sec(x^5)\tan(x^5)5x4sec(x5)tan(x5). Chain rule: multiply by derivative of inner function 5x45x^45x4.

Flashcard 21: What is the derivative of y=cot⁡(ex)y = \cot(e^x)y=cot(ex)?

Answer: −excsc⁡2(ex)-e^x\csc^2(e^x)−excsc2(ex). Chain rule: multiply by derivative of exe^xex which is exe^xex.

Flashcard 22: Find the derivative of f(x)=tan⁡(ex)f(x) = \tan(e^x)f(x)=tan(ex).

Answer: exsec⁡2(ex)e^x\sec^2(e^x)exsec2(ex). Chain rule: multiply by derivative of exe^xex which is exe^xex.

Flashcard 23: Determine the derivative of y=csc⁡(x2)y = \csc(x^2)y=csc(x2).

Answer: −2xcsc⁡(x2)cot⁡(x2)-2x\csc(x^2)\cot(x^2)−2xcsc(x2)cot(x2). Chain rule: multiply by derivative of inner function 2x2x2x.

Flashcard 24: Compute the derivative of h(x)=sec⁡(ln⁡(x))h(x) = \sec(\ln(x))h(x)=sec(ln(x)).

Answer: 1xsec⁡(ln⁡(x))tan⁡(ln⁡(x))\frac{1}{x}\sec(\ln(x))\tan(\ln(x))x1​sec(ln(x))tan(ln(x)). Chain rule: multiply by derivative of ln⁡(x)\ln(x)ln(x) which is 1x\frac{1}{x}x1​.

Flashcard 25: Determine the derivative of y=csc⁡(4x2)y = \csc(4x^2)y=csc(4x2).

Answer: −8xcsc⁡(4x2)cot⁡(4x2)-8x\csc(4x^2)\cot(4x^2)−8xcsc(4x2)cot(4x2). Chain rule: multiply by derivative of inner function 8x8x8x.

Flashcard 26: Find d/dxd/dxd/dx for f(x)=sec⁡(sin⁡(x))f(x) = \sec(\sin(x))f(x)=sec(sin(x)).

Answer: cos⁡(x)sec⁡(sin⁡(x))tan⁡(sin⁡(x))\cos(x)\sec(\sin(x))\tan(\sin(x))cos(x)sec(sin(x))tan(sin(x)). Chain rule: multiply by derivative of sin⁡(x)\sin(x)sin(x) which is cos⁡(x)\cos(x)cos(x).

Flashcard 27: Compute the derivative of g(x)=sec⁡(5x)g(x) = \sec(5x)g(x)=sec(5x).

Answer: 5sec⁡(5x)tan⁡(5x)5\sec(5x)\tan(5x)5sec(5x)tan(5x). Chain rule: multiply by derivative of inner function 555.

Flashcard 28: Find d/dxd/dxd/dx for f(x)=sec⁡2(x)f(x) = \sec^2(x)f(x)=sec2(x).

Answer: 2sec⁡(x)sec⁡(x)tan⁡(x)2\sec(x)\sec(x)\tan(x)2sec(x)sec(x)tan(x). Chain rule applied to sec⁡2(x)=2sec⁡(x)⋅sec⁡(x)tan⁡(x)\sec^2(x) = 2\sec(x) \cdot \sec(x)\tan(x)sec2(x)=2sec(x)⋅sec(x)tan(x).

Flashcard 29: What is the derivative of y=cot⁡(x3)y = \cot(x^3)y=cot(x3)?

Answer: −3x2csc⁡2(x3)-3x^2\csc^2(x^3)−3x2csc2(x3). Chain rule: multiply by derivative of inner function 3x23x^23x2.

Flashcard 30: Determine d/dxd/dxd/dx for h(x)=csc⁡(2x)h(x) = \csc(2x)h(x)=csc(2x).

Answer: −2csc⁡(2x)cot⁡(2x)-2\csc(2x)\cot(2x)−2csc(2x)cot(2x). Chain rule: multiply by derivative of inner function 222.