AP Calculus AB Flashcards: Differentiating Inverse Functions

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 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.

Flashcard 1: Determine if $y = \text{arcsin}(x)$ is differentiable at $x = 1$.

Answer: No, $y = \text{arcsin}(x)$ is not differentiable at $x = 1$. At domain boundary, derivative is undefined.

Flashcard 2: What is the derivative of $y = \text{arctan}(x)$?

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

Flashcard 3: Determine if $y = \text{arcsec}(x)$ is differentiable at $x = 0.5$.

Answer: No, $y = \text{arcsec}(x)$ is not differentiable at $x = 0.5$. Domain of $\arcsec$ is $|x| \geq 1$, so $x = 0.5$ is outside domain.

Flashcard 4: Find the derivative of $y = \text{arcsin}(2x)$ at $x = 0$.

Answer: $2$. Using chain rule: $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}$, at $x=0$ gives $2$.

Flashcard 5: Find the derivative of $y = \text{arctan}(3x)$ at $x = 0$.

Answer: $3$. Using chain rule: $\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+(3x)^2}$, at $x=0$ gives $3$.

Flashcard 6: Find the derivative of $y = \text{arccos}(5x)$ at $x = 0$.

Answer: $-5$. Using chain rule: $\frac{d}{dx}[\arccos(5x)] = -\frac{5}{\sqrt{1-(5x)^2}}$, at $x=0$ gives $-5$.

Flashcard 7: State the formula for the derivative of an inverse function.

Answer: $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$. The inverse function derivative theorem.

Flashcard 8: What is the derivative of $y = \text{arccot}(x)$?

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

Flashcard 9: Find $(f^{-1})'(b)$ if $f(x) = x^3 + x$ and $f(a) = b$.

Answer: $\frac{1}{3a^2+1}$. Using inverse function theorem: $(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{3a^2+1}$.

Flashcard 10: Which inverse function has a derivative of $-\frac{1}{1+x^2}$?

Answer: $\text{arccot}(x)$. Inverse cotangent has derivative $-\frac{1}{1+x^2}$.

Flashcard 11: Find $(f^{-1})'(2)$ if $f(x) = e^x$ and $f(a) = 2$.

Answer: $\frac{1}{2}$. Since $f(\ln 2) = 2$ and $f'(\ln 2) = 2$, so $(f^{-1})'(2) = \frac{1}{2}$.

Flashcard 12: Which inverse function has a derivative of $\frac{1}{1+x^2}$?

Answer: $\text{arctan}(x)$. Inverse tangent has derivative $\frac{1}{1+x^2}$.

Flashcard 13: What is the domain of $y = \text{arccos}(x)$?

Answer: $[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.

Flashcard 14: What is the domain of $y = \text{arcsin}(x)$?

Answer: $[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.

Flashcard 15: Find the derivative of $y = \arccot(6x)$ at $x = 0$.

Answer: $-6$. Using chain rule: $\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}$, at $x=0$ gives $-6$.

Flashcard 16: Find the derivative of $y = \text{arctan}(3x)$ at $x = 0$.

Answer: $3$. Using chain rule: $\frac{d}{dx}[\arctan(3x)] = \frac{3}{1+(3x)^2}$, at $x=0$ gives $3$.

Flashcard 17: Find the derivative of $y = \text{arccos}(5x)$ at $x = 0$.

Answer: $-5$. Using chain rule: $\frac{d}{dx}[\arccos(5x)] = -\frac{5}{\sqrt{1-(5x)^2}}$, at $x=0$ gives $-5$.

Flashcard 18: Find the derivative of $y = \text{arccot}(6x)$ at $x = 0$.

Answer: $-6$. Using chain rule: $\frac{d}{dx}[\arccot(6x)] = -\frac{6}{1+(6x)^2}$, at $x=0$ gives $-6$.

Flashcard 19: Find $(f^{-1})'(b)$ if $f(x) = x^3 + x$ and $f(a) = b$.

Answer: $\frac{1}{3a^2+1}$. Using inverse function theorem: $(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{3a^2+1}$.

Flashcard 20: Determine if $y = \text{arcsin}(x)$ is differentiable at $x = 1$.

Answer: No, $y = \text{arcsin}(x)$ is not differentiable at $x = 1$. At domain boundary, derivative is undefined.

Flashcard 21: Determine if $y = \text{arcsec}(x)$ is differentiable at $x = 0.5$.

Answer: No, $y = \text{arcsec}(x)$ is not differentiable at $x = 0.5$. Domain of $\arcsec$ is $|x| \geq 1$, so $x = 0.5$ is outside domain.

Flashcard 22: What is the domain of $y = \text{arcsin}(x)$?

Answer: $[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.

Flashcard 23: What is the domain of $y = \text{arccos}(x)$?

Answer: $[-1, 1]$. Domain restricted to values where $-1 \leq x \leq 1$.

Flashcard 24: Find the derivative of $y = \text{arcsin}(2x)$ at $x = 0$.

Answer: $2$. Using chain rule: $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-(2x)^2}}$, at $x=0$ gives $2$.

Flashcard 25: State the formula for the derivative of an inverse function.

Answer: $(f^{-1})'(b) = \frac{1}{f'(a)}$ where $f(a) = b$. The inverse function derivative theorem.

Flashcard 26: What is the derivative of $y = \text{arctan}(x)$?

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

Flashcard 27: What is the derivative of $y = \text{arccot}(x)$?

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

Flashcard 28: Which inverse function has a derivative of $-\frac{1}{1+x^2}$?

Answer: $\text{arccot}(x)$. Inverse cotangent has derivative $-\frac{1}{1+x^2}$.

Flashcard 29: Find $(f^{-1})'(2)$ if $f(x) = e^x$ and $f(a) = 2$.

Answer: $\frac{1}{2}$. Since $f(\ln 2) = 2$ and $f'(\ln 2) = 2$, so $(f^{-1})'(2) = \frac{1}{2}$.

Flashcard 30: Which inverse function has a derivative of $\frac{1}{1+x^2}$?

Answer: $\text{arctan}(x)$. Inverse tangent has derivative $\frac{1}{1+x^2}$.