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AP Calculus BC Flashcards: Differentiating Inverse Functions

Study Differentiating Inverse 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

Differentiating Inverse Functions

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QUESTION
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Find (f−1)′(9)(f^{-1})'(9) given f(3)=9f(3)=9 and f′(3)=0f'(3)=0.

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ANSWER

Undefined because f′(3)=0f'(3)=0. Cannot divide by zero when f′(3)=0f'(3)=0.

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What this deck covers

This deck focuses on Differentiating Inverse 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.

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Flashcard 1: Find (f−1)′(9)(f^{-1})'(9) given f(3)=9f(3)=9 and f′(3)=0f'(3)=0.

Answer: Undefined because f′(3)=0f'(3)=0. Cannot divide by zero when f′(3)=0f'(3)=0.

Flashcard 2: What is the geometric meaning of inverse derivatives: how do the tangent line slopes compare at inverse points?

Answer: Slopes are reciprocals at (a,b)(a,b) and (b,a)(b,a). Tangent lines at (a,b)(a,b) and (b,a)(b,a) have reciprocal slopes.

Flashcard 3: Find (f−1)′(4)(f^{-1})'(4) given f(2)=4f(2)=4 and f′(2)=5f'(2)=5.

Answer: (f−1)′(4)=15(f^{-1})'(4)=\frac{1}{5}. Since f(2)=4f(2)=4 and f′(2)=5f'(2)=5, use reciprocal formula.

Flashcard 4: Identify the derivative of y=arctan(x)y = \text{arctan}(x).

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

Flashcard 5: Find (f−1)′(2)(f^{-1})'(2) if f(x)=x2+1f(x)=x^2+1 with domain [0,∞)[0,\infty).

Answer: (f−1)′(2)=12(f^{-1})'(2)=\frac{1}{2}. Since f′(x)=2xf'(x)=2x and f(1)=2f(1)=2, so f′(1)=2f'(1)=2, reciprocal is 12\frac{1}{2}.

Flashcard 6: Find (f−1)′(e)(f^{-1})'(e) if f(x)=ln⁡(x)f(x)=\ln(x).

Answer: (f−1)′(e)=e(f^{-1})'(e)=e. Since exe^x and ln⁡(x)\ln(x) are inverses, f′(1)=1ef'(1)=\frac{1}{e} gives reciprocal ee.

Flashcard 7: Find (f−1)′(π2)(f^{-1})'(\frac{\pi}{2}) if f(x)=sin⁡(x)f(x)=\sin(x) with domain [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right].

Answer: Undefined because cos⁡(π2)=0\cos\left(\frac{\pi}{2}\right)=0. At x=π2x=\frac{\pi}{2}, sin⁡′(x)=cos⁡(x)=0\sin'(x)=\cos(x)=0, so undefined.

Flashcard 8: Find (f−1)′(1)(f^{-1})'(1) if f(x)=exf(x)=e^x.

Answer: (f−1)′(1)=1(f^{-1})'(1)=1. Since ln⁡′(e)=1e\ln'(e)=\frac{1}{e} and ln⁡(e)=1\ln(e)=1, reciprocal is ee.

Flashcard 9: Identify the graph transformation for inverses: across which line are y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) reflected?

Answer: Reflection across y=xy=x. Inverse functions are mirror images across the diagonal line.

Flashcard 10: Identify the key condition needed to differentiate an inverse: what must be true about f′(a)f'(a)?

Answer: f′(a)≠0f'(a)\ne 0. Inverse derivative exists only when original derivative is nonzero.

Flashcard 11: Find (f−1)′(1)(f^{-1})'(1) given f(0)=1f(0)=1 and f′(0)=14f'(0)=\frac{1}{4}.

Answer: (f−1)′(1)=4(f^{-1})'(1)=4. Since f(0)=1f(0)=1 and f′(0)=14f'(0)=\frac{1}{4}, reciprocal is 44.

Flashcard 12: What is the derivative of y=ln∣x∣y = \text{ln}|x|?

Answer: 1x\frac{1}{x} for x≠0x \neq 0. Chain rule applied to absolute value inside logarithm.

Flashcard 13: State the derivative formula for an inverse function: what is (f−1)′(x)(f^{-1})'(x) in terms of f′f'?

Answer: (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}. Derivative of inverse is reciprocal of original derivative at corresponding point.

Flashcard 14: Find (f−1)′(0)(f^{-1})'(0) if f(x)=x+sin⁡(x)f(x)=x+\sin(x) and f(0)=0f(0)=0.

Answer: (f−1)′(0)=12(f^{-1})'(0)=\frac{1}{2}. Since f′(0)=1+cos⁡(0)=2f'(0)=1+\cos(0)=2, reciprocal is 12\frac{1}{2}.

Flashcard 15: Find (f−1)′(8)(f^{-1})'(8) if f(x)=x3f(x)=x^3 and f−1(8)=2f^{-1}(8)=2.

Answer: (f−1)′(8)=112(f^{-1})'(8)=\frac{1}{12}. Since f′(x)=3x2f'(x)=3x^2 and f′(2)=12f'(2)=12, reciprocal is 112\frac{1}{12}.

Flashcard 16: Identify the derivative of y=arccot(x)y = \text{arccot}(x).

Answer: −11+x2-\frac{1}{1+x^2}. Derivative of inverse cotangent, negative of arctan derivative.

Flashcard 17: Identify the derivative of y=arctan(x)y = \text{arctan}(x).

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

Flashcard 18: What is the inverse-derivative relationship between slopes: how are f′(a)f'(a) and (f−1)′(b)(f^{-1})'(b) related when f(a)=bf(a)=b?

Answer: f′(a)⋅(f−1)′(b)=1f'(a)\cdot (f^{-1})'(b)=1. Slopes of inverse functions multiply to 1 at corresponding points.

Flashcard 19: What is the derivative of y=ln(x)y = \text{ln}(x)?

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

Flashcard 20: Find (f−1)′(2)(f^{-1})'(2) if f(x)=x3+1f(x)=x^3+1 and f−1(2)=1f^{-1}(2)=1.

Answer: (f−1)′(2)=13(f^{-1})'(2)=\frac{1}{3}. Since f′(x)=3x2f'(x)=3x^2 and f′(1)=3f'(1)=3, reciprocal is 13\frac{1}{3}.

Flashcard 21: Which theorem relates derivatives of inverses to the original function?

Answer: Inverse Function Theorem. States that (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} when f′(f−1(x))≠0f'(f^{-1}(x)) \neq 0.

Flashcard 22: What is the derivative of y=ln(x)y = \text{ln}(x)?

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

Flashcard 23: Identify the correct expression: which equals (f−1)′(x)(f^{-1})'(x), 1f′(f−1(x))\frac{1}{f'(f^{-1}(x))} or 1f′(x)\frac{1}{f'(x)}?

Answer: 1f′(f−1(x))\frac{1}{f'(f^{-1}(x))}. Must evaluate derivative at f−1(x)f^{-1}(x), not at xx directly.

Flashcard 24: What is the derivative of y=ln∣x∣y = \text{ln}|x|?

Answer: 1x\frac{1}{x} for x≠0x \neq 0. Chain rule applied to absolute value inside logarithm.

Flashcard 25: State the chain-rule identity used for inverse differentiation: what is f(f−1(x))f(f^{-1}(x))?

Answer: f(f−1(x))=xf(f^{-1}(x))=x. Composing a function with its inverse yields the identity function.

Flashcard 26: State the derivative formula for the inverse function of f(x)f(x).

Answer: 1f′(f−1(x))\frac{1}{f'(f^{-1}(x))}. Apply the Inverse Function Theorem: (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}.

Flashcard 27: Identify the derivative of y=arccot(x)y = \text{arccot}(x).

Answer: −11+x2-\frac{1}{1+x^2}. Derivative of inverse cotangent, negative of arctan derivative.

Flashcard 28: State the derivative formula for the inverse function of f(x)f(x).

Answer: 1f′(f−1(x))\frac{1}{f'(f^{-1}(x))}. Apply the Inverse Function Theorem: (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}.

Flashcard 29: What is the equivalent point-slope form for inverses: if f(a)=bf(a)=b, what is (f−1)′(b)(f^{-1})'(b)?

Answer: (f−1)′(b)=1f′(a)(f^{-1})'(b)=\frac{1}{f'(a)}. When f(a)=bf(a)=b, slopes at (a,b)(a,b) and (b,a)(b,a) are reciprocals.

Flashcard 30: Find (f−1)′(0)(f^{-1})'(0) given f(−3)=0f(-3)=0 and f′(−3)=−2f'(-3)=-2.

Answer: (f−1)′(0)=−12(f^{-1})'(0)=-\frac{1}{2}. Since f(−3)=0f(-3)=0 and f′(−3)=−2f'(-3)=-2, take reciprocal.

Flashcard 31: Which theorem relates derivatives of inverses to the original function?

Answer: Inverse Function Theorem. States that (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} when f′(f−1(x))≠0f'(f^{-1}(x)) \neq 0.

Flashcard 32: State the chain-rule identity used for inverse differentiation: what is f−1(f(x))f^{-1}(f(x)) (on the domain of invertibility)?

Answer: f−1(f(x))=xf^{-1}(f(x))=x. Inverse followed by function returns original input on valid domain.