AP Calculus AB Flashcards: Implicit Differentiation

What this deck covers

This deck focuses on Implicit Differentiation, 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: Differentiate $\frac{x}{y} = 1$ with respect to $x$.

Answer: $\frac{y - x\frac{dy}{dx}}{y^2} = 0$. Apply quotient rule: $\frac{d}{dx}[\frac{x}{y}] = \frac{y - x\frac{dy}{dx}}{y^2}$.

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

Answer: $2x + 2y\frac{dy}{dx} = 0$. Apply power rule to each term, using chain rule for $y^2$.

Flashcard 3: Find $\frac{dy}{dx}$ for $y^3 + xy = 1$.

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

Flashcard 4: What is the derivative of $xy = 1$ using implicit differentiation?

Answer: $x\frac{dy}{dx} + y = 0$. Use product rule on $xy$.

Flashcard 5: Find $\frac{dy}{dx}$ for $y = x^2y + x$.

Answer: $\frac{dy}{dx} = 2xy + x^2\frac{dy}{dx} + 1$. Use product rule for $x^2y$ term.

Flashcard 6: What does $\frac{dy}{dx}$ represent in implicit differentiation?

Answer: The derivative of $y$ with respect to $x$. It's the rate of change of $y$ with respect to $x$.

Flashcard 7: Differentiate $x^2 - y^2 = 4$ using implicit differentiation.

Answer: $2x - 2y\frac{dy}{dx} = 0$. Apply power rule to both $x^2$ and $y^2$ terms.

Flashcard 8: What is the first step in implicit differentiation?

Answer: Differentiate both sides with respect to $x$. This sets up the differentiation process.

Flashcard 9: Find the derivative of $xy = x + y$ using implicit differentiation.

Answer: $y + x\frac{dy}{dx} = 1 + \frac{dy}{dx}$. Apply product rule to left side.

Flashcard 10: Differentiate $3x^2 + 4y^2 = 12$ with respect to $x$.

Answer: $6x + 8y\frac{dy}{dx} = 0$. Apply power rule to each term.

Flashcard 11: Differentiate $x^2 + 2xy + y^2 = 16$ with respect to $x$.

Answer: $2x + 2y + 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. This is $(x+y)^2 = 16$; use chain rule.

Flashcard 12: Differentiate $x^2 - y^2 = 4$ using implicit differentiation.

Answer: $2x - 2y\frac{dy}{dx} = 0$. Apply power rule to both $x^2$ and $y^2$ terms.

Flashcard 13: Which rule is applied when differentiating $y^2$ with respect to $x$?

Answer: The Chain Rule. Since $y$ depends on $x$, we need $\frac{dy}{dx}$ when differentiating $y^2$.

Flashcard 14: Differentiate $x^2y + y^3 = 0$ using implicit differentiation.

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

Flashcard 15: Find the derivative of $y$ in $3x^2 + 2xy - y^3 = 0$.

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

Flashcard 16: Differentiate: $x^2 + y^2 = 1$ with respect to $x$.

Answer: $2x + 2y\frac{dy}{dx} = 0$. Apply power rule to each term, using chain rule for $y^2$.

Flashcard 17: Differentiate $x^3 + 3xy + y^3 = 0$ using implicit differentiation.

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

Flashcard 18: Find $\frac{dy}{dx}$ for $y = x^2y + x$.

Answer: $\frac{dy}{dx} = 2xy + x^2\frac{dy}{dx} + 1$. Use product rule for $x^2y$ term.

Flashcard 19: Differentiate $2x^2y = 3$ using implicit differentiation.

Answer: $4xy + 2x^2\frac{dy}{dx} = 0$. Use product rule on $2x^2y$.

Flashcard 20: What is the first step in implicit differentiation?

Answer: Differentiate both sides with respect to $x$. This sets up the differentiation process.

Flashcard 21: Differentiate $x^2 + y^2 = 9$ using implicit differentiation.

Answer: $2x + 2y\frac{dy}{dx} = 0$. Same form as circle equation with radius 3.

Flashcard 22: Differentiate $x^3 + y^3 = 6xy$ using implicit differentiation.

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

Flashcard 23: Differentiate $\frac{x^3}{y} = 1$ with respect to $x$.

Answer: $3x^2 - \frac{x^3\frac{dy}{dx}}{y^2} = 0$. Apply quotient rule to $\frac{x^3}{y}$.

Flashcard 24: Differentiate: $\frac{dy}{dx}$ if $x^2 + y^2 = r^2$.

Answer: $2x + 2y \frac{dy}{dx} = 0$. Same as differentiating $x^2 + y^2 = 1$ but with constant $r^2$.

Flashcard 25: Differentiate: $\frac{dy}{dx}$ if $x^2 + y^2 = r^2$.

Answer: $2x + 2y \frac{dy}{dx} = 0$. Same as differentiating $x^2 + y^2 = 1$ but with constant $r^2$.

Flashcard 26: Differentiate $3x^2 + 4y^2 = 12$ with respect to $x$.

Answer: $6x + 8y \frac{dy}{dx} = 0$. Apply power rule to each term.

Flashcard 27: Differentiate $x^3 + 3xy + y^3 = 0$ using implicit differentiation.

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

Flashcard 28: Differentiate $2x^2y = 3$ using implicit differentiation.

Answer: $4xy + 2x^2\frac{dy}{dx} = 0$. Use product rule on $2x^2y$.

Flashcard 29: Differentiate $x^2 + y^2 = 9$ using implicit differentiation.

Answer: $2x + 2y\frac{dy}{dx} = 0$. Same form as circle equation with radius 3.

Flashcard 30: Differentiate $x^2 + 2xy + y^2 = 16$ with respect to $x$.

Answer: $2x + 2y + 2x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. This is $(x+y)^2 = 16$; use chain rule.

Flashcard 31: Differentiate $xy^2 = 4$ with respect to $x$.

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

Flashcard 32: What is the result of differentiating $x + y = 1$ implicitly?

Answer: $1 + \frac{dy}{dx} = 0$. Each term differentiates to its coefficient.

Flashcard 33: Find $\frac{dy}{dx}$ for $y^3 + xy = 1$.

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

Flashcard 34: Find the derivative of $xy = x + y$ using implicit differentiation.

Answer: $y + x\frac{dy}{dx} = 1 + \frac{dy}{dx}$. Apply product rule to left side.

Flashcard 35: Differentiate $x^2y^2 = 4$ with respect to $x$.

Answer: $2xy^2 + 2x^2y\frac{dy}{dx} = 0$. Use product rule: $\frac{d}{dx}[x^2y^2] = 2x(y^2) + x^2(2y\frac{dy}{dx})$.

Flashcard 36: Differentiate $x^2y + y^3 = 0$ using implicit differentiation.

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

Flashcard 37: Differentiate $xy^2 = 4$ with respect to $x$.

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

Flashcard 38: Find the derivative of $y$ in $3x^2 + 2xy - y^3 = 0$.

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

Flashcard 39: Which rule is applied when differentiating $y^2$ with respect to $x$?

Answer: The Chain Rule. Since $y$ depends on $x$, we need $\frac{dy}{dx}$ when differentiating $y^2$.

Flashcard 40: Find the derivative of $x^2y + y^2 = 10$ using implicit differentiation.

Answer: $2xy + x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. Use product rule for $x^2y$ and power rule for $y^2$.

Flashcard 41: Find the derivative of $x^2 + xy + y^2 = 7$.

Answer: $2x + y + x \frac{dy}{dx} + 2y \frac{dy}{dx} = 0$. Apply product rule to $xy$ and power rule to other terms.

Flashcard 42: Differentiate $x^3 + y^3 = 6xy$ using implicit differentiation.

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

Flashcard 43: Differentiate $\frac{x}{y} = 1$ with respect to $x$.

Answer: $\frac{y - x\frac{dy}{dx}}{y^2} = 0$. Apply quotient rule: $\frac{d}{dx}[\frac{x}{y}] = \frac{y - x\frac{dy}{dx}}{y^2}$.

Flashcard 44: What is implicit differentiation?

Answer: A technique to find derivatives of equations not solved for $y$. Used when $y$ is not isolated on one side of the equation.

Flashcard 45: Differentiate $\frac{x^3}{y} = 1$ with respect to $x$.

Answer: $3x^2 - \frac{x^3\frac{dy}{dx}}{y^2} = 0$. Apply quotient rule to $\frac{x^3}{y}$.

Flashcard 46: What is the result of differentiating $x + y = 1$ implicitly?

Answer: $1 + \frac{dy}{dx} = 0$. Each term differentiates to its coefficient.

Flashcard 47: What is the derivative of $xy = 1$ using implicit differentiation?

Answer: $x\frac{dy}{dx} + y = 0$. Use product rule on $xy$.

Flashcard 48: Find $\frac{dy}{dx}$ for $y^2 = x^2 + 2$.

Answer: $2y\frac{dy}{dx} = 2x$. Apply power rule to $y^2$ using chain rule.

Flashcard 49: Find the derivative of $x^2 + xy + y^2 = 7$.

Answer: $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. Apply product rule to $xy$ and power rule to other terms.

Flashcard 50: Differentiate $x^2y^2 = 4$ with respect to $x$.

Answer: $2xy^2 + 2x^2y\frac{dy}{dx} = 0$. Use product rule: $\frac{d}{dx}[x^2y^2] = 2x(y^2) + x^2(2y\frac{dy}{dx})$.

Flashcard 51: Find $\frac{dy}{dx}$ for $y^2 = x^2 + 2$.

Answer: $2y\frac{dy}{dx} = 2x$. Apply power rule to $y^2$ using chain rule.

Flashcard 52: Differentiate $x + y = xy$ with respect to $x$.

Answer: $1 + \frac{dy}{dx} = y + x\frac{dy}{dx}$. Use product rule on right side $xy$.

Flashcard 53: Find the derivative of $x^2y + y^2 = 10$ using implicit differentiation.

Answer: $2xy + x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. Use product rule for $x^2y$ and power rule for $y^2$.

Flashcard 54: Differentiate $x + y = xy$ with respect to $x$.

Answer: $1 + \frac{dy}{dx} = y + x\frac{dy}{dx}$. Use product rule on right side $xy$.