AP Calculus AB Flashcards: Exploring Behaviors Of Implicit Relations

What this deck covers

This deck focuses on Exploring Behaviors Of Implicit Relations, 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: Find $\frac{dy}{dx}$ for $y = \ln(xy)$ implicitly.

Answer: $\frac{dy}{dx} = \frac{1}{x + y}$. From $\frac{dy}{dx} = \frac{1}{xy}(y + x\frac{dy}{dx})$, solve for $\frac{dy}{dx}$.

Flashcard 2: Differentiate $x^2 - y^2 = 1$ implicitly.

Answer: $2x - 2y \frac{dy}{dx} = 0$. Difference of squares: coefficients have opposite signs.

Flashcard 3: Differentiate $x^3 + y^3 = 3xy$ implicitly.

Answer: $3x^2 + 3y^2 \frac{dy}{dx} = 3(y + x \frac{dy}{dx})$. Apply product rule to right side.

Flashcard 4: Differentiate $x^2y^2 = 1$ implicitly.

Answer: $2xy^2 + 2x^2y \frac{dy}{dx} = 0$. Product rule: $2xy^2 + 2x^2y\frac{dy}{dx} = 0$.

Flashcard 5: Differentiate $x + y = e^{xy}$ implicitly.

Answer: $1 + \frac{dy}{dx} = e^{xy}(y + x \frac{dy}{dx})$. Chain rule on exponential function on right side.

Flashcard 6: Differentiate $y = x^2 + \tan(y)$ implicitly.

Answer: $\frac{dy}{dx} = 2x + \sec^2(y) \frac{dy}{dx}$. Chain rule on $\tan(y)$ gives $\sec^2(y)\frac{dy}{dx}$.

Flashcard 7: Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$.

Answer: $\frac{dy}{dx} = \frac{y - 2x}{2y - x}$. Differentiate each term, collect $\frac{dy}{dx}$ terms, solve.

Flashcard 8: Differentiate $x^4 + y^4 = 16$ implicitly.

Answer: $4x^3 + 4y^3 \frac{dy}{dx} = 0$. Apply power rule: $4x^3 + 4y^3\frac{dy}{dx} = 0$.

Flashcard 9: Differentiate $x^2 + y^2 = 1$ implicitly with respect to $x$.

Answer: $2x + 2y \frac{dy}{dx} = 0$. Apply power rule to both terms, treat $y$ as function of $x$.

Flashcard 10: What is the implicit differentiation of $x^2y + y^2 = 1$?

Answer: $2xy + x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$. Product rule on $x^2y$, power rule on $y^2$.

Flashcard 11: What is the implicit differential of $x^2 + xy = 10$?

Answer: $2x + y + x \frac{dy}{dx} = 0$. Product rule on $xy$ term.

Flashcard 12: What is the implicit derivative of $y^3 + 3x^2y = 12$?

Answer: $3y^2 \frac{dy}{dx} + 6xy + 3x^2 \frac{dy}{dx} = 0$. Apply chain rule to $y^3$ and product rule to $3x^2y$.

Flashcard 13: Differentiate $y^2 = x + x^2y$ implicitly.

Answer: $2y \frac{dy}{dx} = 1 + 2xy + x^2 \frac{dy}{dx}$. Product rule on right side, power rule on left.

Flashcard 14: Differentiate $\sin(x + y) = y$ implicitly.

Answer: $\cos(x + y)(1 + \frac{dy}{dx}) = \frac{dy}{dx}$. Chain rule on $\sin(x + y)$ gives $\cos(x + y)(1 + \frac{dy}{dx})$.

Flashcard 15: What does $\frac{dy}{dx}$ represent in implicit differentiation?

Answer: The derivative of $y$ with respect to $x$. The rate of change of $y$ with respect to $x$.

Flashcard 16: What is the implicit derivative of $y = \sin(xy)$?

Answer: $\frac{dy}{dx} = \cos(xy)(y + x \frac{dy}{dx})$. Chain rule on $\sin(xy)$ equals $\frac{dy}{dx}$.

Flashcard 17: Differentiate $x = \cos(xy)$ implicitly.

Answer: $1 = -\sin(xy)(y + x \frac{dy}{dx})$. Derivative of $\cos$ is $-\sin$, apply chain rule.

Flashcard 18: Find $\frac{dy}{dx}$ for $y = \cos(xy)$ implicitly.

Answer: $\frac{dy}{dx} = -\sin(xy)(y + x \frac{dy}{dx})$. Chain rule: $\frac{dy}{dx} = -\sin(xy)(y + x\frac{dy}{dx})$.

Flashcard 19: What is the implicit derivative of $y = x + \sin(y)$?

Answer: $\frac{dy}{dx} = 1 + \cos(y) \frac{dy}{dx}$. Chain rule on $\sin(y)$ gives $\cos(y)\frac{dy}{dx}$.

Flashcard 20: Find $\frac{dy}{dx}$ for $xy + y = 3x$.

Answer: $\frac{dy}{dx} = \frac{3 - y}{x + 1}$. Factor out terms with $y$: $y(x + 1) = 3x$.

Flashcard 21: Differentiate $x = \tan(xy)$ implicitly.

Answer: $1 = \sec^2(xy)(y + x \frac{dy}{dx})$. Derivative of $\tan$ is $\sec^2$, apply chain rule.

Flashcard 22: Find $\frac{dy}{dx}$ for $xy = \ln(x)$.

Answer: $\frac{dy}{dx} = \frac{1 - y}{x}$. From product rule: $y + x\frac{dy}{dx} = \frac{1}{x}$.

Flashcard 23: Differentiate $x^2 + y^2 = 25$ implicitly.

Answer: $2x + 2y \frac{dy}{dx} = 0$. Same as $x^2 + y^2 = 1$ with different radius.

Flashcard 24: What is the implicit derivative of $x^3 + y^3 = 6xy$?

Answer: $3x^2 + 3y^2 \frac{dy}{dx} = 6(y + x \frac{dy}{dx})$. Apply product rule to right side: $6(y + x\frac{dy}{dx})$.

Flashcard 25: What is implicit differentiation?

Answer: Differentiating equations not solved for one variable in terms of others. Used when $y$ cannot be easily isolated.

Flashcard 26: State the Chain Rule for implicit differentiation.

Answer: Differentiate outer, multiply by derivative of inner. Essential for composite functions involving $y$.

Flashcard 27: What is the purpose of implicit differentiation?

Answer: To find derivatives when not in explicit form. Enables differentiation of relations not solved for $y$.

Flashcard 28: Differentiate $e^{xy} = x + y$ implicitly with respect to $x$.

Answer: $e^{xy}(y + x \frac{dy}{dx}) = 1 + \frac{dy}{dx}$. Chain rule on left, standard derivatives on right.

Flashcard 29: Differentiate $\tan(xy) = x$ implicitly.

Answer: $\sec^2(xy)(y + x \frac{dy}{dx}) = 1$. Chain rule: derivative of $\tan$ is $\sec^2$.

Flashcard 30: Find $\frac{dy}{dx}$ for $\cos(xy) = x$.

Answer: $\frac{dy}{dx} = \frac{y\sin(xy) - 1}{x\sin(xy)}$. Chain rule gives $-\sin(xy)(y + x\frac{dy}{dx}) = 1$, solve.