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AP Calculus BC Flashcards: Differentiating Inverse Trigonometric Functions

Study Differentiating Inverse Trigonometric 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 Differentiating Inverse Trigonometric 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: Differentiating Inverse Trigonometric Functions

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QUESTION

Evaluate ddx(tan⁡−1(x))\frac{d}{dx}(\tan^{-1}(x))dxd​(tan−1(x)) at x=1x = 1x=1.

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ANSWER

12\frac{1}{2}21​. Substitute x=1x = 1x=1 into 11+x2\frac{1}{1+x^2}1+x21​.

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Flashcard 1: Evaluate ddx(tan⁡−1(x))\frac{d}{dx}(\tan^{-1}(x))dxd​(tan−1(x)) at x=1x = 1x=1.

Answer: 12\frac{1}{2}21​. Substitute x=1x = 1x=1 into 11+x2\frac{1}{1+x^2}1+x21​.

Flashcard 2: What is the chain rule application for y=cot⁡−1(g(x))y = \cot^{-1}(g(x))y=cot−1(g(x))?

Answer: −g′(x)1+(g(x))2-\frac{g'(x)}{1+(g(x))^2}−1+(g(x))2g′(x)​. General chain rule pattern for inverse cotangent composition.

Flashcard 3: Identify the derivative of y=cot⁡−1(x)y = \cot^{-1}(\sqrt{x})y=cot−1(x​).

Answer: −12x(1+x)-\frac{1}{2\sqrt{x}(1+x)}−2x​(1+x)1​. Chain rule with ddx(x)=12x\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}dxd​(x​)=2x​1​ applied to cot⁡−1\cot^{-1}cot−1.

Flashcard 4: Identify the derivative of y=cos⁡−1(x3)y = \cos^{-1}(x^3)y=cos−1(x3).

Answer: −3x21−x6-\frac{3x^2}{\sqrt{1-x^6}}−1−x6​3x2​. Chain rule with ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2dxd​(x3)=3x2 applied to cos⁡−1\cos^{-1}cos−1 formula.

Flashcard 5: Evaluate ddx(sin⁡−1(x))\frac{d}{dx}(\sin^{-1}(x))dxd​(sin−1(x)) at x=12x = \frac{1}{2}x=21​.

Answer: 233\frac{2\sqrt{3}}{3}323​​. Substitute x=12x = \frac{1}{2}x=21​ into 11−x2\frac{1}{\sqrt{1-x^2}}1−x2​1​.

Flashcard 6: Find ddx(cot⁡−1(2x))\frac{d}{dx}(\cot^{-1}(\sqrt{2x}))dxd​(cot−1(2x​)).

Answer: −122x(1+2x)-\frac{1}{2\sqrt{2x}(1+2x)}−22x​(1+2x)1​. Chain rule with ddx(2x)=12x\frac{d}{dx}(\sqrt{2x}) = \frac{1}{\sqrt{2x}}dxd​(2x​)=2x​1​ applied to cot⁡−1\cot^{-1}cot−1.

Flashcard 7: State the derivative of sec⁡−1(x)\sec^{-1}(x)sec−1(x).

Answer: 1∣x∣x2−1\frac{1}{|x|\sqrt{x^2-1}}∣x∣x2−1​1​. Basic derivative formula for inverse secant function.

Flashcard 8: State the derivative of tan⁡−1(x)\tan^{-1}(x)tan−1(x).

Answer: 11+x2\frac{1}{1+x^2}1+x21​. Basic derivative formula for inverse tangent function.

Flashcard 9: Identify the derivative of y=csc⁡−1(x5)y = \csc^{-1}(x^5)y=csc−1(x5).

Answer: −5x4∣x5∣x10−1-\frac{5x^4}{|x^5|\sqrt{x^{10}-1}}−∣x5∣x10−1​5x4​. Chain rule with ddx(x5)=5x4\frac{d}{dx}(x^5) = 5x^4dxd​(x5)=5x4 applied to csc⁡−1\csc^{-1}csc−1 formula.

Flashcard 10: What is the chain rule application for y=sin⁡−1(g(x))y = \sin^{-1}(g(x))y=sin−1(g(x))?

Answer: g′(x)1−(g(x))2\frac{g'(x)}{\sqrt{1-(g(x))^2}}1−(g(x))2​g′(x)​. General chain rule pattern for inverse sine composition.

Flashcard 11: Find ddx(sin⁡−1(2x))\frac{d}{dx}(\sin^{-1}(2x))dxd​(sin−1(2x)).

Answer: 21−(2x)2\frac{2}{\sqrt{1-(2x)^2}}1−(2x)2​2​. Chain rule with ddx(2x)=2\frac{d}{dx}(2x) = 2dxd​(2x)=2 applied to sin⁡−1\sin^{-1}sin−1 formula.

Flashcard 12: What is the chain rule application for y=csc⁡−1(g(x))y = \csc^{-1}(g(x))y=csc−1(g(x))?

Answer: −g′(x)∣g(x)∣(g(x))2−1-\frac{g'(x)}{|g(x)|\sqrt{(g(x))^2-1}}−∣g(x)∣(g(x))2−1​g′(x)​. General chain rule pattern for inverse cosecant composition.

Flashcard 13: State the derivative of csc⁡−1(x)\csc^{-1}(x)csc−1(x).

Answer: −1∣x∣x2−1-\frac{1}{|x|\sqrt{x^2-1}}−∣x∣x2−1​1​. Basic derivative formula for inverse cosecant function.

Flashcard 14: Find ddx(tan⁡−1(3x))\frac{d}{dx}(\tan^{-1}(\sqrt{3x}))dxd​(tan−1(3x​)).

Answer: 323x(1+3x)\frac{3}{2\sqrt{3x}(1+3x)}23x​(1+3x)3​. Chain rule with ddx(3x)=323x\frac{d}{dx}(\sqrt{3x}) = \frac{3}{2\sqrt{3x}}dxd​(3x​)=23x​3​ applied to tan⁡−1\tan^{-1}tan−1.

Flashcard 15: Find ddx(csc⁡−1(2x3))\frac{d}{dx}(\csc^{-1}(2x^3))dxd​(csc−1(2x3)).

Answer: −6x2∣2x3∣(2x3)2−1-\frac{6x^2}{|2x^3|\sqrt{(2x^3)^2-1}}−∣2x3∣(2x3)2−1​6x2​. Chain rule with ddx(2x3)=6x2\frac{d}{dx}(2x^3) = 6x^2dxd​(2x3)=6x2 applied to csc⁡−1\csc^{-1}csc−1.

Flashcard 16: Find ddx(sec⁡−1(5x2))\frac{d}{dx}(\sec^{-1}(5x^2))dxd​(sec−1(5x2)).

Answer: 10x∣5x2∣(5x2)2−1\frac{10x}{|5x^2|\sqrt{(5x^2)^2-1}}∣5x2∣(5x2)2−1​10x​. Chain rule with ddx(5x2)=10x\frac{d}{dx}(5x^2) = 10xdxd​(5x2)=10x applied to sec⁡−1\sec^{-1}sec−1.

Flashcard 17: Find ddx(cot⁡−1(4x))\frac{d}{dx}(\cot^{-1}(4x))dxd​(cot−1(4x)).

Answer: −41+(4x)2-\frac{4}{1+(4x)^2}−1+(4x)24​. Chain rule with ddx(4x)=4\frac{d}{dx}(4x) = 4dxd​(4x)=4 applied to cot⁡−1\cot^{-1}cot−1 formula.

Flashcard 18: State the derivative of cos⁡−1(x)\cos^{-1}(x)cos−1(x).

Answer: −11−x2-\frac{1}{\sqrt{1-x^2}}−1−x2​1​. Basic derivative formula for inverse cosine function.

Flashcard 19: Find ddx(xcos⁡−1(x))\frac{d}{dx}(x \cos^{-1}(x))dxd​(xcos−1(x)).

Answer: cos⁡−1(x)−x1−x2\cos^{-1}(x) - \frac{x}{\sqrt{1-x^2}}cos−1(x)−1−x2​x​. Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′ applied to xcos⁡−1(x)x\cos^{-1}(x)xcos−1(x).

Flashcard 20: Evaluate ddx(csc⁡−1(x))\frac{d}{dx}(\csc^{-1}(x))dxd​(csc−1(x)) at x=2x = 2x=2.

Answer: −123-\frac{1}{2\sqrt{3}}−23​1​. Substitute x=2x = 2x=2 into −1∣x∣x2−1-\frac{1}{|x|\sqrt{x^2-1}}−∣x∣x2−1​1​.

Flashcard 21: Evaluate ddx(sec⁡−1(x))\frac{d}{dx}(\sec^{-1}(x))dxd​(sec−1(x)) at x=2x = 2x=2.

Answer: 123\frac{1}{2\sqrt{3}}23​1​. Substitute x=2x = 2x=2 into 1∣x∣x2−1\frac{1}{|x|\sqrt{x^2-1}}∣x∣x2−1​1​.

Flashcard 22: Find ddx(sin⁡−1(3x2))\frac{d}{dx}(\sin^{-1}(3x^2))dxd​(sin−1(3x2)).

Answer: 6x1−(3x2)2\frac{6x}{\sqrt{1-(3x^2)^2}}1−(3x2)2​6x​. Chain rule with ddx(3x2)=6x\frac{d}{dx}(3x^2) = 6xdxd​(3x2)=6x applied to sin⁡−1\sin^{-1}sin−1.

Flashcard 23: What is the chain rule application for y=tan⁡−1(g(x))y = \tan^{-1}(g(x))y=tan−1(g(x))?

Answer: g′(x)1+(g(x))2\frac{g'(x)}{1+(g(x))^2}1+(g(x))2g′(x)​. General chain rule pattern for inverse tangent composition.

Flashcard 24: Find ddx(tan⁡−1(5x))\frac{d}{dx}(\tan^{-1}(5x))dxd​(tan−1(5x)).

Answer: 51+(5x)2\frac{5}{1+(5x)^2}1+(5x)25​. Chain rule with ddx(5x)=5\frac{d}{dx}(5x) = 5dxd​(5x)=5 applied to tan⁡−1\tan^{-1}tan−1 formula.

Flashcard 25: What is the chain rule application for y=sec⁡−1(g(x))y = \sec^{-1}(g(x))y=sec−1(g(x))?

Answer: g′(x)∣g(x)∣(g(x))2−1\frac{g'(x)}{|g(x)|\sqrt{(g(x))^2-1}}∣g(x)∣(g(x))2−1​g′(x)​. General chain rule pattern for inverse secant composition.

Flashcard 26: Identify the derivative of y=sin⁡−1(x2)y = \sin^{-1}(x^2)y=sin−1(x2).

Answer: 2x1−x4\frac{2x}{\sqrt{1-x^4}}1−x4​2x​. Chain rule with ddx(x2)=2x\frac{d}{dx}(x^2) = 2xdxd​(x2)=2x applied to sin⁡−1\sin^{-1}sin−1 formula.

Flashcard 27: Find ddx(csc⁡−1(7x))\frac{d}{dx}(\csc^{-1}(7x))dxd​(csc−1(7x)).

Answer: −7∣7x∣(7x)2−1-\frac{7}{|7x|\sqrt{(7x)^2-1}}−∣7x∣(7x)2−1​7​. Chain rule with ddx(7x)=7\frac{d}{dx}(7x) = 7dxd​(7x)=7 applied to csc⁡−1\csc^{-1}csc−1 formula.

Flashcard 28: Find ddx(sec⁡−1(6x))\frac{d}{dx}(\sec^{-1}(6x))dxd​(sec−1(6x)).

Answer: 6∣6x∣(6x)2−1\frac{6}{|6x|\sqrt{(6x)^2-1}}∣6x∣(6x)2−1​6​. Chain rule with ddx(6x)=6\frac{d}{dx}(6x) = 6dxd​(6x)=6 applied to sec⁡−1\sec^{-1}sec−1 formula.

Flashcard 29: State the derivative of sin⁡−1(x)\sin^{-1}(x)sin−1(x).

Answer: 11−x2\frac{1}{\sqrt{1-x^2}}1−x2​1​. Basic derivative formula for inverse sine function.

Flashcard 30: Find ddx(cos⁡−1(3x))\frac{d}{dx}(\cos^{-1}(3x))dxd​(cos−1(3x)).

Answer: −31−(3x)2-\frac{3}{\sqrt{1-(3x)^2}}−1−(3x)2​3​. Chain rule with ddx(3x)=3\frac{d}{dx}(3x) = 3dxd​(3x)=3 applied to cos⁡−1\cos^{-1}cos−1 formula.