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AP Calculus BC Flashcards: Implicit Differentiation

Study Implicit Differentiation in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

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.

AP Calculus BC Flashcards: Implicit Differentiation

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QUESTION

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

Tap or drag to reveal answer

ANSWER

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

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: Find $\frac{dy}{dx}$ for $xy = 1$ .

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

Flashcard 8: Differentiate $xy = 1$ implicitly.

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

Flashcard 9: 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 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: 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 27: 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 28: 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 29: 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 30: 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 $\frac{1}{x+y} = (x+y)^{-1}$ .

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AP Calculus BC Flashcards: Implicit Differentiation

Study Implicit Differentiation in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

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.

AP Calculus BC Flashcards: Implicit Differentiation

/ 30

0 reviewed

0% Complete

0 reviewing

QUESTION

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

Tap or drag to reveal answer

ANSWER

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

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: Find $\frac{dy}{dx}$ for $xy = 1$ .

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

Flashcard 8: Differentiate $xy = 1$ implicitly.

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

Flashcard 9: 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 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: 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 27: 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 28: 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 29: 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 30: 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 $\frac{1}{x+y} = (x+y)^{-1}$ .

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AP Calculus BC Flashcards: Implicit Differentiation

What this deck covers

How to use these flashcards

AP Calculus BC Flashcards: Implicit Differentiation

All flashcards

Flashcard 1: Differentiate cos(x+y)=x\text{cos}(x + y) = xcos(x+y)=x implicitly.

Flashcard 2: Differentiate y3+3xy=6y^3 + 3xy = 6y3+3xy=6 implicitly.

Flashcard 3: Differentiate y2+yx=1y^2 + yx = 1y2+yx=1 implicitly.

Flashcard 4: Differentiate x2y+y2x=1x^2y + y^2x = 1x2y+y2x=1 implicitly.

Flashcard 5: Find dydx\frac{dy}{dx}dxdy​ for x2+y2=1x^2 + y^2 = 1x2+y2=1.

Flashcard 6: Differentiate xy=1\frac{x}{y} = 1yx​=1 implicitly.

Flashcard 7: Find dydx\frac{dy}{dx}dxdy​ for xy=1xy = 1xy=1.

Flashcard 8: Differentiate xy=1xy = 1xy=1 implicitly.

Flashcard 9: Find dydx\frac{dy}{dx}dxdy​ for x2−xy+y2=7x^2 - xy + y^2 = 7x2−xy+y2=7.

Flashcard 10: Differentiate ln(x+y)=x−y\text{ln}(x + y) = x - yln(x+y)=x−y implicitly.

Flashcard 11: Differentiate ex+y=xye^{x+y} = xyex+y=xy implicitly.

Flashcard 12: Differentiate x2+y2=4xyx^2 + y^2 = 4xyx2+y2=4xy implicitly.

Flashcard 13: State the chain rule for differentiation.

Flashcard 14: Differentiate exy=ye^{xy} = yexy=y implicitly.

Flashcard 15: Differentiate x2+y2=1x^2 + y^2 = 1x2+y2=1 implicitly.

Flashcard 16: Find dydx\frac{dy}{dx}dxdy​ for 1x+y=x−y\frac{1}{x+y} = x-yx+y1​=x−y.

Flashcard 17: Find dydx\frac{dy}{dx}dxdy​ for x2y+y2x=1x^2y + y^2x = 1x2y+y2x=1.

Flashcard 18: Differentiate sin(xy)=y\text{sin}(xy) = ysin(xy)=y implicitly.

Flashcard 19: Find dydx\frac{dy}{dx}dxdy​ for y2+yx=1y^2 + yx = 1y2+yx=1.

Flashcard 20: Find dydx\frac{dy}{dx}dxdy​ for x2+2y2=3xyx^2 + 2y^2 = 3xyx2+2y2=3xy.

Flashcard 21: Differentiate x3+y3=6xyx^3 + y^3 = 6xyx3+y3=6xy implicitly.

Flashcard 22: Find dydx\frac{dy}{dx}dxdy​ for xy=1\frac{x}{y} = 1yx​=1.

Flashcard 23: Find dydx\frac{dy}{dx}dxdy​ for ln(x+y)=x−y\text{ln}(x + y) = x - yln(x+y)=x−y.

Flashcard 24: Differentiate 1x+y=x−y\frac{1}{x+y} = x-yx+y1​=x−y implicitly.

Flashcard 25: Differentiate x2+2y2=3xyx^2 + 2y^2 = 3xyx2+2y2=3xy implicitly.

Flashcard 26: Differentiate ln(xy)=y2\text{ln}(xy) = y^2ln(xy)=y2 implicitly.

Flashcard 27: Find dydx\frac{dy}{dx}dxdy​ for ex+y=xye^{x+y} = xyex+y=xy.

Flashcard 28: Differentiate x2−xy+y2=7x^2 - xy + y^2 = 7x2−xy+y2=7 implicitly.

Flashcard 29: Find dydx\frac{dy}{dx}dxdy​ for y2+yx=1y^2 + yx = 1y2+yx=1.

Flashcard 30: Differentiate 1x+y=x−y\frac{1}{x+y} = x-yx+y1​=x−y implicitly.

Opening subject page...

AP Calculus BC Flashcards: Implicit Differentiation

What this deck covers

How to use these flashcards

AP Calculus BC Flashcards: Implicit Differentiation

All flashcards

Flashcard 1: Differentiate cos(x+y)=x\text{cos}(x + y) = xcos(x+y)=x implicitly.

Flashcard 2: Differentiate y3+3xy=6y^3 + 3xy = 6y3+3xy=6 implicitly.

Flashcard 3: Differentiate y2+yx=1y^2 + yx = 1y2+yx=1 implicitly.

Flashcard 4: Differentiate x2y+y2x=1x^2y + y^2x = 1x2y+y2x=1 implicitly.

Flashcard 5: Find dydx\frac{dy}{dx}dxdy​ for x2+y2=1x^2 + y^2 = 1x2+y2=1.

Flashcard 6: Differentiate xy=1\frac{x}{y} = 1yx​=1 implicitly.

Flashcard 7: Find dydx\frac{dy}{dx}dxdy​ for xy=1xy = 1xy=1.

Flashcard 8: Differentiate xy=1xy = 1xy=1 implicitly.

Flashcard 9: Find dydx\frac{dy}{dx}dxdy​ for x2−xy+y2=7x^2 - xy + y^2 = 7x2−xy+y2=7.

Flashcard 10: Differentiate ln(x+y)=x−y\text{ln}(x + y) = x - yln(x+y)=x−y implicitly.

Flashcard 11: Differentiate ex+y=xye^{x+y} = xyex+y=xy implicitly.

Flashcard 12: Differentiate x2+y2=4xyx^2 + y^2 = 4xyx2+y2=4xy implicitly.

Flashcard 13: State the chain rule for differentiation.

Flashcard 14: Differentiate exy=ye^{xy} = yexy=y implicitly.

Flashcard 15: Differentiate x2+y2=1x^2 + y^2 = 1x2+y2=1 implicitly.

Flashcard 16: Find dydx\frac{dy}{dx}dxdy​ for 1x+y=x−y\frac{1}{x+y} = x-yx+y1​=x−y.

Flashcard 17: Find dydx\frac{dy}{dx}dxdy​ for x2y+y2x=1x^2y + y^2x = 1x2y+y2x=1.

Flashcard 18: Differentiate sin(xy)=y\text{sin}(xy) = ysin(xy)=y implicitly.

Flashcard 19: Find dydx\frac{dy}{dx}dxdy​ for y2+yx=1y^2 + yx = 1y2+yx=1.

Flashcard 20: Find dydx\frac{dy}{dx}dxdy​ for x2+2y2=3xyx^2 + 2y^2 = 3xyx2+2y2=3xy.

Flashcard 21: Differentiate x3+y3=6xyx^3 + y^3 = 6xyx3+y3=6xy implicitly.

Flashcard 22: Find dydx\frac{dy}{dx}dxdy​ for xy=1\frac{x}{y} = 1yx​=1.

Flashcard 23: Find dydx\frac{dy}{dx}dxdy​ for ln(x+y)=x−y\text{ln}(x + y) = x - yln(x+y)=x−y.

Flashcard 24: Differentiate 1x+y=x−y\frac{1}{x+y} = x-yx+y1​=x−y implicitly.

Flashcard 25: Differentiate x2+2y2=3xyx^2 + 2y^2 = 3xyx2+2y2=3xy implicitly.

Flashcard 26: Differentiate ln(xy)=y2\text{ln}(xy) = y^2ln(xy)=y2 implicitly.

Flashcard 27: Find dydx\frac{dy}{dx}dxdy​ for ex+y=xye^{x+y} = xyex+y=xy.

Flashcard 28: Differentiate x2−xy+y2=7x^2 - xy + y^2 = 7x2−xy+y2=7 implicitly.

Flashcard 29: Find dydx\frac{dy}{dx}dxdy​ for y2+yx=1y^2 + yx = 1y2+yx=1.

Flashcard 30: Differentiate 1x+y=x−y\frac{1}{x+y} = x-yx+y1​=x−y implicitly.

Flashcard 1: Differentiate $\text{cos}(x + y) = x$ implicitly.

Flashcard 2: Differentiate $y^3 + 3xy = 6$ implicitly.

Flashcard 3: Differentiate $y^2 + yx = 1$ implicitly.

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

Flashcard 5: Find $\frac{dy}{dx}$ for $x^2 + y^2 = 1$ .

Flashcard 6: Differentiate $\frac{x}{y} = 1$ implicitly.

Flashcard 7: Find $\frac{dy}{dx}$ for $xy = 1$ .

Flashcard 8: Differentiate $xy = 1$ implicitly.

Flashcard 9: Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$ .

Flashcard 10: Differentiate $\text{ln}(x + y) = x - y$ implicitly.

Flashcard 11: Differentiate $e^{x+y} = xy$ implicitly.

Flashcard 12: Differentiate $x^2 + y^2 = 4xy$ implicitly.

Flashcard 14: Differentiate $e^{xy} = y$ implicitly.

Flashcard 15: Differentiate $x^2 + y^2 = 1$ implicitly.

Flashcard 16: Find $\frac{dy}{dx}$ for $\frac{1}{x+y} = x-y$ .

Flashcard 17: Find $\frac{dy}{dx}$ for $x^2y + y^2x = 1$ .

Flashcard 18: Differentiate $\text{sin}(xy) = y$ implicitly.

Flashcard 19: Find $\frac{dy}{dx}$ for $y^2 + yx = 1$ .

Flashcard 20: Find $\frac{dy}{dx}$ for $x^2 + 2y^2 = 3xy$ .

Flashcard 21: Differentiate $x^3 + y^3 = 6xy$ implicitly.

Flashcard 22: Find $\frac{dy}{dx}$ for $\frac{x}{y} = 1$ .

Flashcard 23: Find $\frac{dy}{dx}$ for $\text{ln}(x + y) = x - y$ .

Flashcard 24: Differentiate $\frac{1}{x+y} = x-y$ implicitly.

Flashcard 25: Differentiate $x^2 + 2y^2 = 3xy$ implicitly.

Flashcard 26: Differentiate $\text{ln}(xy) = y^2$ implicitly.

Flashcard 27: Find $\frac{dy}{dx}$ for $e^{x+y} = xy$ .

Flashcard 28: Differentiate $x^2 - xy + y^2 = 7$ implicitly.

Flashcard 29: Find $\frac{dy}{dx}$ for $y^2 + yx = 1$ .

Flashcard 30: Differentiate $\frac{1}{x+y} = x-y$ implicitly.

Flashcard 1: Differentiate $\text{cos}(x + y) = x$ implicitly.

Flashcard 2: Differentiate $y^3 + 3xy = 6$ implicitly.

Flashcard 3: Differentiate $y^2 + yx = 1$ implicitly.

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

Flashcard 5: Find $\frac{dy}{dx}$ for $x^2 + y^2 = 1$ .

Flashcard 6: Differentiate $\frac{x}{y} = 1$ implicitly.

Flashcard 7: Find $\frac{dy}{dx}$ for $xy = 1$ .

Flashcard 8: Differentiate $xy = 1$ implicitly.

Flashcard 9: Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 7$ .

Flashcard 10: Differentiate $\text{ln}(x + y) = x - y$ implicitly.

Flashcard 11: Differentiate $e^{x+y} = xy$ implicitly.

Flashcard 12: Differentiate $x^2 + y^2 = 4xy$ implicitly.

Flashcard 14: Differentiate $e^{xy} = y$ implicitly.

Flashcard 15: Differentiate $x^2 + y^2 = 1$ implicitly.

Flashcard 16: Find $\frac{dy}{dx}$ for $\frac{1}{x+y} = x-y$ .

Flashcard 17: Find $\frac{dy}{dx}$ for $x^2y + y^2x = 1$ .

Flashcard 18: Differentiate $\text{sin}(xy) = y$ implicitly.

Flashcard 19: Find $\frac{dy}{dx}$ for $y^2 + yx = 1$ .

Flashcard 20: Find $\frac{dy}{dx}$ for $x^2 + 2y^2 = 3xy$ .

Flashcard 21: Differentiate $x^3 + y^3 = 6xy$ implicitly.

Flashcard 22: Find $\frac{dy}{dx}$ for $\frac{x}{y} = 1$ .

Flashcard 23: Find $\frac{dy}{dx}$ for $\text{ln}(x + y) = x - y$ .

Flashcard 24: Differentiate $\frac{1}{x+y} = x-y$ implicitly.

Flashcard 25: Differentiate $x^2 + 2y^2 = 3xy$ implicitly.

Flashcard 26: Differentiate $\text{ln}(xy) = y^2$ implicitly.

Flashcard 27: Find $\frac{dy}{dx}$ for $e^{x+y} = xy$ .

Flashcard 28: Differentiate $x^2 - xy + y^2 = 7$ implicitly.

Flashcard 29: Find $\frac{dy}{dx}$ for $y^2 + yx = 1$ .

Flashcard 30: Differentiate $\frac{1}{x+y} = x-y$ implicitly.