AP Calculus BC Flashcards: Differentiating Inverse Trigonometric Functions

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.

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

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

Flashcard 2: Find $\frac{d}{dx}(\sin^{-1}(2x))$.

Answer: $\frac{2}{\sqrt{1-(2x)^2}}$. Chain rule with $\frac{d}{dx}(2x) = 2$ applied to $\sin^{-1}$ formula.

Flashcard 3: Identify the derivative of $y = \csc^{-1}(x^5)$.

Answer: $-\frac{5x^4}{|x^5|\sqrt{x^{10}-1}}$. Chain rule with $\frac{d}{dx}(x^5) = 5x^4$ applied to $\csc^{-1}$ formula.

Flashcard 4: Find $\frac{d}{dx}(\sin^{-1}(3x^2))$.

Answer: $\frac{6x}{\sqrt{1-(3x^2)^2}}$. Chain rule with $\frac{d}{dx}(3x^2) = 6x$ applied to $\sin^{-1}$.

Flashcard 5: Find $\frac{d}{dx}(\sin^{-1}(2x))$.

Answer: $\frac{2}{\sqrt{1-(2x)^2}}$. Chain rule with $\frac{d}{dx}(2x) = 2$ applied to $\sin^{-1}$ formula.

Flashcard 6: Find $\frac{d}{dx}(\cos^{-1}(4x^3))$.

Answer: $-\frac{12x^2}{\sqrt{1-(4x^3)^2}}$. Chain rule with $\frac{d}{dx}(4x^3) = 12x^2$ applied to $\cos^{-1}$.

Flashcard 7: Find $\frac{d}{dx}(\tan^{-1}(\sqrt{3x}))$.

Answer: $\frac{3}{2\sqrt{3x}(1+3x)}$. Chain rule with $\frac{d}{dx}(\sqrt{3x}) = \frac{3}{2\sqrt{3x}}$ applied to $\tan^{-1}$.

Flashcard 8: Evaluate $\frac{d}{dx}(\csc^{-1}(x))$ at $x = 2$.

Answer: $-\frac{1}{2\sqrt{3}}$. Substitute $x = 2$ into $-\frac{1}{|x|\sqrt{x^2-1}}$.

Flashcard 9: Evaluate $\frac{d}{dx}(\cot^{-1}(x))$ at $x = 1$.

Answer: $-\frac{1}{2}$. Substitute $x = 1$ into $-\frac{1}{1+x^2}$.

Flashcard 10: Find $\frac{d}{dx}(\cot^{-1}(\sqrt{2x}))$.

Answer: $-\frac{1}{2\sqrt{2x}(1+2x)}$. Chain rule with $\frac{d}{dx}(\sqrt{2x}) = \frac{1}{\sqrt{2x}}$ applied to $\cot^{-1}$.

Flashcard 11: Find $\frac{d}{dx}(\sec^{-1}(5x^2))$.

Answer: $\frac{10x}{|5x^2|\sqrt{(5x^2)^2-1}}$. Chain rule with $\frac{d}{dx}(5x^2) = 10x$ applied to $\sec^{-1}$.

Flashcard 12: Find $\frac{d}{dx}(\cot^{-1}(\sqrt{2x}))$.

Answer: $-\frac{1}{2\sqrt{2x}(1+2x)}$. Chain rule with $\frac{d}{dx}(\sqrt{2x}) = \frac{1}{\sqrt{2x}}$ applied to $\cot^{-1}$.

Flashcard 13: State the derivative of $\sin^{-1}(x)$.

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

Flashcard 14: State the derivative of $\csc^{-1}(x)$.

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

Flashcard 15: Identify the derivative of $y = \cos^{-1}(x^3)$.

Answer: $-\frac{3x^2}{\sqrt{1-x^6}}$. Chain rule with $\frac{d}{dx}(x^3) = 3x^2$ applied to $\cos^{-1}$ formula.

Flashcard 16: What is the chain rule application for $y = \cos^{-1}(g(x))$?

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

Flashcard 17: Evaluate $\frac{d}{dx}(\cot^{-1}(x))$ at $x = 1$.

Answer: $-\frac{1}{2}$. Substitute $x = 1$ into $-\frac{1}{1+x^2}$.

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

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

Flashcard 19: Identify the derivative of $y = \cot^{-1}(\sqrt{x})$.

Answer: $-\frac{1}{2\sqrt{x}(1+x)}$. Chain rule with $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$ applied to $\cot^{-1}$.

Flashcard 20: Find $\frac{d}{dx}(\csc^{-1}(2x^3))$.

Answer: $-\frac{6x^2}{|2x^3|\sqrt{(2x^3)^2-1}}$. Chain rule with $\frac{d}{dx}(2x^3) = 6x^2$ applied to $\csc^{-1}$.

Flashcard 21: What is the chain rule application for $y = \cos^{-1}(g(x))$?

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

Flashcard 22: State the derivative of $\cot^{-1}(x)$.

Answer: $-\frac{1}{1+x^2}$. Basic derivative formula for inverse cotangent function.

Flashcard 23: Identify the derivative of $y = \tan^{-1}(\sqrt{x})$.

Answer: $\frac{1}{2\sqrt{x}(1+x)}$. Chain rule with $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$ applied to $\tan^{-1}$.

Flashcard 24: Evaluate $\frac{d}{dx}(\csc^{-1}(x))$ at $x = 2$.

Answer: $-\frac{1}{2\sqrt{3}}$. Substitute $x = 2$ into $-\frac{1}{|x|\sqrt{x^2-1}}$.

Flashcard 25: Find $\frac{d}{dx}(\cos^{-1}(4x^3))$.

Answer: $-\frac{12x^2}{\sqrt{1-(4x^3)^2}}$. Chain rule with $\frac{d}{dx}(4x^3) = 12x^2$ applied to $\cos^{-1}$.

Flashcard 26: Identify the derivative of $y = \sec^{-1}(x^4)$.

Answer: $\frac{4x^3}{|x^4|\sqrt{x^8-1}}$. Chain rule with $\frac{d}{dx}(x^4) = 4x^3$ applied to $\sec^{-1}$ formula.

Flashcard 27: Find $\frac{d}{dx}(\cot^{-1}(4x))$.

Answer: $-\frac{4}{1+(4x)^2}$. Chain rule with $\frac{d}{dx}(4x) = 4$ applied to $\cot^{-1}$ formula.

Flashcard 28: Evaluate $\frac{d}{dx}(\sin^{-1}(x))$ at $x = \frac{1}{2}$.

Answer: $\frac{2\sqrt{3}}{3}$. Substitute $x = \frac{1}{2}$ into $\frac{1}{\sqrt{1-x^2}}$.

Flashcard 29: Identify the derivative of $y = \tan^{-1}(\sqrt{x})$.

Answer: $\frac{1}{2\sqrt{x}(1+x)}$. Chain rule with $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$ applied to $\tan^{-1}$.

Flashcard 30: Identify the derivative of $y = \cos^{-1}(x^3)$.

Answer: $-\frac{3x^2}{\sqrt{1-x^6}}$. Chain rule with $\frac{d}{dx}(x^3) = 3x^2$ applied to $\cos^{-1}$ formula.

Flashcard 31: State the derivative of $\cos^{-1}(x)$.

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

Flashcard 32: Find $\frac{d}{dx}(\sec^{-1}(6x))$.

Answer: $\frac{6}{|6x|\sqrt{(6x)^2-1}}$. Chain rule with $\frac{d}{dx}(6x) = 6$ applied to $\sec^{-1}$ formula.

Flashcard 33: What is the chain rule application for $y = \cot^{-1}(g(x))$?

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

Flashcard 34: Find $\frac{d}{dx}(x \cos^{-1}(x))$.

Answer: $\cos^{-1}(x) - \frac{x}{\sqrt{1-x^2}}$. Product rule: $(uv)' = u'v + uv'$ applied to $x\cos^{-1}(x)$.

Flashcard 35: Find $\frac{d}{dx}(\cos^{-1}(3x))$.

Answer: $-\frac{3}{\sqrt{1-(3x)^2}}$. Chain rule with $\frac{d}{dx}(3x) = 3$ applied to $\cos^{-1}$ formula.

Flashcard 36: Find $\frac{d}{dx}(\tan^{-1}(5x))$.

Answer: $\frac{5}{1+(5x)^2}$. Chain rule with $\frac{d}{dx}(5x) = 5$ applied to $\tan^{-1}$ formula.

Flashcard 37: Identify the derivative of $y = \sin^{-1}(x^2)$.

Answer: $\frac{2x}{\sqrt{1-x^4}}$. Chain rule with $\frac{d}{dx}(x^2) = 2x$ applied to $\sin^{-1}$ formula.

Flashcard 38: Identify the derivative of $y = \cot^{-1}(\sqrt{x})$.

Answer: $-\frac{1}{2\sqrt{x}(1+x)}$. Chain rule with $\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$ applied to $\cot^{-1}$.

Flashcard 39: Evaluate $\frac{d}{dx}(\tan^{-1}(x))$ at $x = 1$.

Answer: $\frac{1}{2}$. Substitute $x = 1$ into $\frac{1}{1+x^2}$.

Flashcard 40: Evaluate $\frac{d}{dx}(\sin^{-1}(x))$ at $x = \frac{1}{2}$.

Answer: $\frac{2\sqrt{3}}{3}$. Substitute $x = \frac{1}{2}$ into $\frac{1}{\sqrt{1-x^2}}$.

Flashcard 41: What is the chain rule application for $y = \csc^{-1}(g(x))$?

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

Flashcard 42: Evaluate $\frac{d}{dx}(\cos^{-1}(x))$ at $x = \frac{1}{2}$.

Answer: $-\frac{2\sqrt{3}}{3}$. Substitute $x = \frac{1}{2}$ into $-\frac{1}{\sqrt{1-x^2}}$.

Flashcard 43: Find $\frac{d}{dx}(x \sin^{-1}(x))$.

Answer: $\sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}}$. Product rule: $(uv)' = u'v + uv'$ applied to $x\sin^{-1}(x)$.

Flashcard 44: Find $\frac{d}{dx}(\sec^{-1}(6x))$.

Answer: $\frac{6}{|6x|\sqrt{(6x)^2-1}}$. Chain rule with $\frac{d}{dx}(6x) = 6$ applied to $\sec^{-1}$ formula.

Flashcard 45: Find $\frac{d}{dx}(x \sin^{-1}(x))$.

Answer: $\sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}}$. Product rule: $(uv)' = u'v + uv'$ applied to $x\sin^{-1}(x)$.

Flashcard 46: State the derivative of $\tan^{-1}(x)$.

Answer: $\frac{1}{1+x^2}$. Basic derivative formula for inverse tangent function.

Flashcard 47: Find $\frac{d}{dx}(\csc^{-1}(2x^3))$.

Answer: $-\frac{6x^2}{|2x^3|\sqrt{(2x^3)^2-1}}$. Chain rule with $\frac{d}{dx}(2x^3) = 6x^2$ applied to $\csc^{-1}$.

Flashcard 48: Find $\frac{d}{dx}(\sec^{-1}(5x^2))$.

Answer: $\frac{10x}{|5x^2|\sqrt{(5x^2)^2-1}}$. Chain rule with $\frac{d}{dx}(5x^2) = 10x$ applied to $\sec^{-1}$.

Flashcard 49: What is the chain rule application for $y = \tan^{-1}(g(x))$?

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

Flashcard 50: Find $\frac{d}{dx}(\csc^{-1}(7x))$.

Answer: $-\frac{7}{|7x|\sqrt{(7x)^2-1}}$. Chain rule with $\frac{d}{dx}(7x) = 7$ applied to $\csc^{-1}$ formula.

Flashcard 51: Identify the derivative of $y = \sec^{-1}(x^4)$.

Answer: $\frac{4x^3}{|x^4|\sqrt{x^8-1}}$. Chain rule with $\frac{d}{dx}(x^4) = 4x^3$ applied to $\sec^{-1}$ formula.

Flashcard 52: State the derivative of $\sin^{-1}(x)$.

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

Flashcard 53: Find $\frac{d}{dx}(\cos^{-1}(3x))$.

Answer: $-\frac{3}{\sqrt{1-(3x)^2}}$. Chain rule with $\frac{d}{dx}(3x) = 3$ applied to $\cos^{-1}$ formula.

Flashcard 54: Identify the derivative of $y = \sin^{-1}(x^2)$.

Answer: $\frac{2x}{\sqrt{1-x^4}}$. Chain rule with $\frac{d}{dx}(x^2) = 2x$ applied to $\sin^{-1}$ formula.

Flashcard 55: State the derivative of $\sec^{-1}(x)$.

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

Flashcard 56: Evaluate $\frac{d}{dx}(\cos^{-1}(x))$ at $x = \frac{1}{2}$.

Answer: $-\frac{2\sqrt{3}}{3}$. Substitute $x = \frac{1}{2}$ into $-\frac{1}{\sqrt{1-x^2}}$.

Flashcard 57: Evaluate $\frac{d}{dx}(\tan^{-1}(x))$ at $x = 1$.

Answer: $\frac{1}{2}$. Substitute $x = 1$ into $\frac{1}{1+x^2}$.

Flashcard 58: Find $\frac{d}{dx}(\sin^{-1}(3x^2))$.

Answer: $\frac{6x}{\sqrt{1-(3x^2)^2}}$. Chain rule with $\frac{d}{dx}(3x^2) = 6x$ applied to $\sin^{-1}$.

Flashcard 59: What is the chain rule application for $y = \csc^{-1}(g(x))$?

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

Flashcard 60: What is the chain rule application for $y = \sec^{-1}(g(x))$?

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

Flashcard 61: Find $\frac{d}{dx}(\tan^{-1}(5x))$.

Answer: $\frac{5}{1+(5x)^2}$. Chain rule with $\frac{d}{dx}(5x) = 5$ applied to $\tan^{-1}$ formula.

Flashcard 62: What is the chain rule application for $y = \cot^{-1}(g(x))$?

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

Flashcard 63: What is the chain rule application for $y = \sin^{-1}(g(x))$?

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

Flashcard 64: State the derivative of $\tan^{-1}(x)$.

Answer: $\frac{1}{1+x^2}$. Basic derivative formula for inverse tangent function.

Flashcard 65: What is the chain rule application for $y = \tan^{-1}(g(x))$?

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

Flashcard 66: Find $\frac{d}{dx}(\cot^{-1}(4x))$.

Answer: $-\frac{4}{1+(4x)^2}$. Chain rule with $\frac{d}{dx}(4x) = 4$ applied to $\cot^{-1}$ formula.

Flashcard 67: Identify the derivative of $y = \csc^{-1}(x^5)$.

Answer: $-\frac{5x^4}{|x^5|\sqrt{x^{10}-1}}$. Chain rule with $\frac{d}{dx}(x^5) = 5x^4$ applied to $\csc^{-1}$ formula.

Flashcard 68: State the derivative of $\cot^{-1}(x)$.

Answer: $-\frac{1}{1+x^2}$. Basic derivative formula for inverse cotangent function.

Flashcard 69: What is the chain rule application for $y = \sin^{-1}(g(x))$?

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

Flashcard 70: State the derivative of $\sec^{-1}(x)$.

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

Flashcard 71: Find $\frac{d}{dx}(\csc^{-1}(7x))$.

Answer: $-\frac{7}{|7x|\sqrt{(7x)^2-1}}$. Chain rule with $\frac{d}{dx}(7x) = 7$ applied to $\csc^{-1}$ formula.

Flashcard 72: Evaluate $\frac{d}{dx}(\sec^{-1}(x))$ at $x = 2$.

Answer: $\frac{1}{2\sqrt{3}}$. Substitute $x = 2$ into $\frac{1}{|x|\sqrt{x^2-1}}$.

Flashcard 73: Find $\frac{d}{dx}(x \cos^{-1}(x))$.

Answer: $\cos^{-1}(x) - \frac{x}{\sqrt{1-x^2}}$. Product rule: $(uv)' = u'v + uv'$ applied to $x\cos^{-1}(x)$.

Flashcard 74: Find $\frac{d}{dx}(\tan^{-1}(\sqrt{3x}))$.

Answer: $\frac{3}{2\sqrt{3x}(1+3x)}$. Chain rule with $\frac{d}{dx}(\sqrt{3x}) = \frac{3}{2\sqrt{3x}}$ applied to $\tan^{-1}$.

Flashcard 75: State the derivative of $\csc^{-1}(x)$.

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

Flashcard 76: Evaluate $\frac{d}{dx}(\sec^{-1}(x))$ at $x = 2$.

Answer: $\frac{1}{2\sqrt{3}}$. Substitute $x = 2$ into $\frac{1}{|x|\sqrt{x^2-1}}$.