AP Calculus BC 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 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: Find $\frac{dy}{dx}$ for $\frac{x}{y} = 1$.

Answer: $\frac{dy}{dx} = \frac{y}{x}$. From $\frac{x}{y} = 1$, we get $x = y$, so slopes are equal.

Flashcard 2: Find $\frac{dy}{dx}$ for $\frac{x}{y} = 1$.

Answer: $\frac{dy}{dx} = \frac{y}{x}$. From $\frac{x}{y} = 1$, we get $x = y$, so slopes are equal.

Flashcard 3: Find $\frac{dy}{dx}$ for $\text{sin}(xy) = x$.

Answer: $\frac{dy}{dx} = \frac{1 - y\text{cos}(xy)}{x\text{cos}(xy)}$. Solve for $\frac{dy}{dx}$ from the differentiated equation.

Flashcard 4: Differentiate $x^3 + y^3 = 6xy$ implicitly.

Answer: $3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$. Apply power rule to each cubic term, product rule to $6xy$.

Flashcard 5: Find $\frac{dy}{dx}$ for $x^2 + 2y^2 = 3xy$.

Answer: $\frac{dy}{dx} = \frac{3y - 2x}{4y - 3x}$. Collect $\frac{dy}{dx}$ terms and solve for the derivative.

Flashcard 6: Differentiate $x^3 + y^3 = 6xy$ implicitly.

Answer: $3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$. Apply power rule to each cubic term, product rule to $6xy$.

Flashcard 7: Differentiate $x^2 + y^2 = 1$ implicitly.

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

Flashcard 8: Differentiate $\text{sin}(xy) = x$ implicitly.

Answer: $\text{cos}(xy)(y + x\frac{dy}{dx}) = 1$. Chain rule: $\frac{d}{dx}[\sin(u)] = \cos(u)\frac{du}{dx}$.

Flashcard 9: Find $\frac{dy}{dx}$ for $e^{x+y} = xy$.

Answer: $\frac{dy}{dx} = \frac{y - e^{x+y}}{e^{x+y} - x}$. Collect $\frac{dy}{dx}$ terms and solve algebraically.

Flashcard 10: Differentiate $x^2y + y^2x = 1$ implicitly.

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

Flashcard 11: Find $\frac{dy}{dx}$ for $x^2y + y^2x = 1$.

Answer: $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2yx}$. Factor out $\frac{dy}{dx}$ from numerator and solve.

Flashcard 12: What is implicit differentiation?

Answer: Differentiating both sides of an equation with respect to $x$. Used when $y$ is not explicitly solved for in terms of $x$.

Flashcard 13: Differentiate $x^2 + y^2 = 4xy$ implicitly.

Answer: $2x + 2y\frac{dy}{dx} = 4y + 4x\frac{dy}{dx}$. Standard implicit differentiation with product rule on right.

Flashcard 14: Differentiate $\text{cos}(x + y) = x$ implicitly.

Answer: $-\text{sin}(x+y)(1 + \frac{dy}{dx}) = 1$. Chain rule: $\frac{d}{dx}[\cos(u)] = -\sin(u)\frac{du}{dx}$.

Flashcard 15: Differentiate $x^2 + 2y^2 = 3xy$ implicitly.

Answer: $2x + 4y\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$. Apply power rule to each term, product rule to $3xy$.

Flashcard 16: Find $\frac{dy}{dx}$ for $\text{sin}(xy) = x$.

Answer: $\frac{dy}{dx} = \frac{1 - y\text{cos}(xy)}{x\text{cos}(xy)}$. Solve for $\frac{dy}{dx}$ from the differentiated equation.

Flashcard 17: Differentiate $x^2 - xy + y^2 = 7$ implicitly.

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

Flashcard 18: Differentiate $x^2 + y^2 = 4xy$ implicitly.

Answer: $2x + 2y\frac{dy}{dx} = 4y + 4x\frac{dy}{dx}$. Standard implicit differentiation with product rule on right.

Flashcard 19: Differentiate $xy = 1$ implicitly.

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

Flashcard 20: Find $\frac{dy}{dx}$ for $\frac{1}{x+y} = x-y$.

Answer: $\frac{dy}{dx} = \frac{(x+y)^2 - 1}{1 + (x+y)^2}$. Solve by collecting $\frac{dy}{dx}$ terms on one side.

Flashcard 21: Find $\frac{dy}{dx}$ for $xy = 1$.

Answer: $\frac{dy}{dx} = -\frac{y}{x}$. Solve for $\frac{dy}{dx}$ from the differentiated equation.

Flashcard 22: Differentiate $\text{sin}(xy) = y$ implicitly.

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

Flashcard 23: Differentiate $x^2 + y^2 = 1$ implicitly.

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

Flashcard 24: Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$.

Answer: $\frac{dy}{dx} = \frac{2x - y}{x - 2y}$. Collect $\frac{dy}{dx}$ terms and solve for the derivative.

Flashcard 25: Find $\frac{dy}{dx}$ for $\ln(x + y) = x - y$.

Answer: $\frac{dy}{dx} = \frac{x+y-1}{x+y+1}$. Collect $\frac{dy}{dx}$ terms and solve algebraically.

Flashcard 26: Find $\frac{dy}{dx}$ for $e^{x+y} = xy$.

Answer: $\frac{dy}{dx} = \frac{y - e^{x+y}}{e^{x+y} - x}$. Collect $\frac{dy}{dx}$ terms and solve algebraically.

Flashcard 27: Differentiate $x^2y + y^2x = 1$ implicitly.

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

Flashcard 28: Differentiate $y^3 + 3xy = 6$ implicitly.

Answer: $3y^2\frac{dy}{dx} + 3y + 3x\frac{dy}{dx} = 0$. Power rule on $y^3$, product rule on $3xy$ term.

Flashcard 29: Differentiate $\text{sin}(xy) = y$ implicitly.

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

Flashcard 30: Find $\frac{dy}{dx}$ for $y^2 + yx = 1$.

Answer: $\frac{dy}{dx} = \frac{-y}{x + 2y}$. Factor $\frac{dy}{dx}$ and solve for it algebraically.

Flashcard 31: State the chain rule for differentiation.

Answer: If $y=f(u)$ and $u=g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. Essential for differentiating composite functions.

Flashcard 32: Differentiate $y^3 + 3xy = 6$ implicitly.

Answer: $3y^2\frac{dy}{dx} + 3y + 3x\frac{dy}{dx} = 0$. Power rule on $y^3$, product rule on $3xy$ term.

Flashcard 33: Differentiate $\frac{x}{y} = 1$ implicitly.

Answer: $\frac{y - x\frac{dy}{dx}}{y^2} = 0$. Use quotient rule: $\frac{d}{dx}[\frac{u}{v}] = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Flashcard 34: Differentiate $\frac{x}{y} = 1$ implicitly.

Answer: $\frac{y - x\frac{dy}{dx}}{y^2} = 0$. Use quotient rule: $\frac{d}{dx}[\frac{u}{v}] = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Flashcard 35: Find $\frac{dy}{dx}$ for $x^2 + y^2 = 1$.

Answer: $\frac{dy}{dx} = -\frac{x}{y}$. Solve for $\frac{dy}{dx}$ by isolating it algebraically.

Flashcard 36: Find $\frac{dy}{dx}$ for $x^2y + y^2x = 1$.

Answer: $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2yx}$. Factor out $\frac{dy}{dx}$ from numerator and solve.

Flashcard 37: Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$.

Answer: $\frac{dy}{dx} = \frac{2x - y}{x - 2y}$. Collect $\frac{dy}{dx}$ terms and solve for the derivative.

Flashcard 38: Differentiate $\text{ln}(xy) = y^2$ implicitly.

Answer: $\frac{1}{xy}(y + x\frac{dy}{dx}) = 2y\frac{dy}{dx}$. Chain rule on $\ln(xy)$ and product rule within.

Flashcard 39: Differentiate $\text{cos}(x + y) = x$ implicitly.

Answer: $-\text{sin}(x+y)(1 + \frac{dy}{dx}) = 1$. Chain rule: $\frac{d}{dx}[\cos(u)] = -\sin(u)\frac{du}{dx}$.

Flashcard 40: Find $\frac{dy}{dx}$ for $\text{ln}(x + y) = x - y$.

Answer: $\frac{dy}{dx} = \frac{x+y-1}{x+y+1}$. Collect $\frac{dy}{dx}$ terms and solve algebraically.

Flashcard 41: Differentiate $xy = 1$ implicitly.

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

Flashcard 42: Differentiate $\text{ln}(x + y) = x - y$ implicitly.

Answer: $\frac{1}{x+y}(1 + \frac{dy}{dx}) = 1 - \frac{dy}{dx}$. Chain rule for $\ln(x+y)$ gives $\frac{1}{x+y}(1 + \frac{dy}{dx})$.

Flashcard 43: Find $\frac{dy}{dx}$ for $\frac{1}{x+y} = x-y$.

Answer: $\frac{dy}{dx} = \frac{(x+y)^2 - 1}{1 + (x+y)^2}$. Solve by collecting $\frac{dy}{dx}$ terms on one side.

Flashcard 44: Differentiate $y^2 + yx = 1$ implicitly.

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

Flashcard 45: Differentiate $x^2 - xy + y^2 = 7$ implicitly.

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

Flashcard 46: Differentiate $\frac{1}{x+y} = x-y$ implicitly.

Answer: $-\frac{1}{(x+y)^2}(1 + \frac{dy}{dx}) = 1 - \frac{dy}{dx}$. Use chain rule on $\frac{1}{x+y} = (x+y)^{-1}$.

Flashcard 47: Differentiate $e^{xy} = y$ implicitly.

Answer: $e^{xy}(y + x\frac{dy}{dx}) = \frac{dy}{dx}$. Chain rule on left, product rule within exponent.

Flashcard 48: Find $\frac{dy}{dx}$ for $y^2 + yx = 1$.

Answer: $\frac{dy}{dx} = \frac{-y}{x + 2y}$. Factor $\frac{dy}{dx}$ and solve for it algebraically.

Flashcard 49: Differentiate $y^2 + yx = 1$ implicitly.

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

Flashcard 50: State the chain rule for differentiation.

Answer: If $y=f(u)$ and $u=g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. Essential for differentiating composite functions.

Flashcard 51: Differentiate $e^{xy} = y$ implicitly.

Answer: $e^{xy}(y + x\frac{dy}{dx}) = \frac{dy}{dx}$. Chain rule on left, product rule within exponent.

Flashcard 52: Differentiate $e^{x+y} = xy$ implicitly.

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

Flashcard 53: Differentiate $e^{x+y} = xy$ implicitly.

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

Flashcard 54: Find $\frac{dy}{dx}$ for $x^2 + y^2 = 1$.

Answer: $\frac{dy}{dx} = -\frac{x}{y}$. Solve for $\frac{dy}{dx}$ by isolating it algebraically.

Flashcard 55: Differentiate $\frac{1}{x+y} = x-y$ implicitly.

Answer: $-\frac{1}{(x+y)^2}(1 + \frac{dy}{dx}) = 1 - \frac{dy}{dx}$. Use chain rule on $\frac{1}{x+y} = (x+y)^{-1}$.

Flashcard 56: Find $\frac{dy}{dx}$ for $xy = 1$.

Answer: $\frac{dy}{dx} = -\frac{y}{x}$. Solve for $\frac{dy}{dx}$ from the differentiated equation.

Flashcard 57: Differentiate $\text{ln}(x + y) = x - y$ implicitly.

Answer: $\frac{1}{x+y}(1 + \frac{dy}{dx}) = 1 - \frac{dy}{dx}$. Chain rule for $\ln(x+y)$ gives $\frac{1}{x+y}(1 + \frac{dy}{dx})$.

Flashcard 58: Differentiate $\text{ln}(xy) = y^2$ implicitly.

Answer: $\frac{1}{xy}(y + x\frac{dy}{dx}) = 2y\frac{dy}{dx}$. Chain rule on $\ln(xy)$ and product rule within.

Flashcard 59: Differentiate $x^2 + 2y^2 = 3xy$ implicitly.

Answer: $2x + 4y\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$. Apply power rule to each term, product rule to $3xy$.

Flashcard 60: Find $\frac{dy}{dx}$ for $x^2 + 2y^2 = 3xy$.

Answer: $\frac{dy}{dx} = \frac{3y - 2x}{4y - 3x}$. Collect $\frac{dy}{dx}$ terms and solve for the derivative.