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AP Calculus AB Quiz

AP Calculus AB Quiz: Exploring Behaviors Of Implicit Relations

Practice Exploring Behaviors Of Implicit Relations in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Given sin⁡(xy)+y=3\sin(xy)+y=3sin(xy)+y=3, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

Select an answer to continue

What this quiz covers

This quiz focuses on Exploring Behaviors Of Implicit Relations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Given sin⁡(xy)+y=3\sin(xy)+y=3sin(xy)+y=3, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. dydx=−cos⁡(xy)\dfrac{dy}{dx}=-\cos(xy)dxdy​=−cos(xy)
  2. dydx=−ycos⁡(xy)xcos⁡(xy)+1\dfrac{dy}{dx}=\dfrac{-y\cos(xy)}{x\cos(xy)+1}dxdy​=xcos(xy)+1−ycos(xy)​ (correct answer)
  3. dydx=−xcos⁡(xy)ycos⁡(xy)+1\dfrac{dy}{dx}=\dfrac{-x\cos(xy)}{y\cos(xy)+1}dxdy​=ycos(xy)+1−xcos(xy)​
  4. dydx=−cos⁡(xy)xcos⁡(xy)+1\dfrac{dy}{dx}=\dfrac{-\cos(xy)}{x\cos(xy)+1}dxdy​=xcos(xy)+1−cos(xy)​
  5. dydx=ycos⁡(xy)xcos⁡(xy)+1\dfrac{dy}{dx}=\dfrac{y\cos(xy)}{x\cos(xy)+1}dxdy​=xcos(xy)+1ycos(xy)​

Explanation: This problem uses implicit differentiation on sin(xy) + y = 3. Differentiating both sides with respect to x gives cos(xy)·(x·dy/dx + y·1) + dy/dx = 0, where we use the chain rule on sin(xy) and the product rule on its argument xy. Expanding yields x·cos(xy)·dy/dx + y·cos(xy) + dy/dx = 0. Factoring out dy/dx gives dy/dx[x·cos(xy) + 1] = -y·cos(xy). Therefore, dy/dx = -y·cos(xy)/[x·cos(xy) + 1]. A tempting error is to forget the chain rule and write dy/dx = -cos(xy)/[x·cos(xy) + 1] (choice D), missing the y factor. Remember: when differentiating composite functions like sin(xy), apply the chain rule first, then handle the inner function.

Question 2

If sin⁡(xy)=x+y\sin(xy)=x+ysin(xy)=x+y, what is dydx\dfrac{dy}{dx}dxdy​ in terms of xxx and yyy?

  1. dydx=cos⁡(xy)(y+xdydx)−11\dfrac{dy}{dx}=\dfrac{\cos(xy)(y+x\dfrac{dy}{dx})-1}{1}dxdy​=1cos(xy)(y+xdxdy​)−1​
  2. dydx=1−cos⁡(xy) ycos⁡(xy) x−1\dfrac{dy}{dx}=\dfrac{1-\cos(xy)\,y}{\cos(xy)\,x-1}dxdy​=cos(xy)x−11−cos(xy)y​ (correct answer)
  3. dydx=cos⁡(xy) y−1cos⁡(xy) x−1\dfrac{dy}{dx}=\dfrac{\cos(xy)\,y-1}{\cos(xy)\,x-1}dxdy​=cos(xy)x−1cos(xy)y−1​
  4. dydx=1−cos⁡(xy) xcos⁡(xy) y−1\dfrac{dy}{dx}=\dfrac{1-\cos(xy)\,x}{\cos(xy)\,y-1}dxdy​=cos(xy)y−11−cos(xy)x​
  5. dydx=cos⁡(xy)1\dfrac{dy}{dx}=\dfrac{\cos(xy)}{1}dxdy​=1cos(xy)​

Explanation: This problem involves implicit differentiation of sin(xy) = x + y, where we must apply the chain rule to a composite function. Differentiating sin(xy) gives cos(xy)·d/dx(xy) = cos(xy)·(y + x(dy/dx)), while the right side gives 1 + dy/dx. The equation becomes cos(xy)·y + cos(xy)·x(dy/dx) = 1 + dy/dx, which rearranges to cos(xy)·x(dy/dx) - dy/dx = 1 - cos(xy)·y. Factoring: dy/dx[cos(xy)·x - 1] = 1 - cos(xy)·y, so dy/dx = (1 - cos(xy)·y)/(cos(xy)·x - 1). Choice C has the wrong sign on the 1 in the numerator, a common error when moving terms across the equation. Remember that when a term moves from left to right (or vice versa), its sign changes.

Question 3

On the level curve ln⁡(x+y)+xy=3\ln(x+y)+xy=3ln(x+y)+xy=3, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. dydx=−1x+y+y1x+y+x\dfrac{dy}{dx}=-\dfrac{\frac{1}{x+y}+y}{\frac{1}{x+y}+x}dxdy​=−x+y1​+xx+y1​+y​ (correct answer)
  2. dydx=−1x+y+x1x+y+y\dfrac{dy}{dx}=-\dfrac{\frac{1}{x+y}+x}{\frac{1}{x+y}+y}dxdy​=−x+y1​+yx+y1​+x​
  3. dydx=−1x+y1x+y\dfrac{dy}{dx}=-\dfrac{\frac{1}{x+y}}{\frac{1}{x+y}}dxdy​=−x+y1​x+y1​​
  4. dydx=−1x+y+y1x+y\dfrac{dy}{dx}=-\dfrac{\frac{1}{x+y}+y}{\frac{1}{x+y}}dxdy​=−x+y1​x+y1​+y​
  5. dydx=−yx\dfrac{dy}{dx}=-\dfrac{y}{x}dxdy​=−xy​

Explanation: This problem requires implicit differentiation to find dy/dx for the level curve ln(x + y) + xy = 3. Differentiating gives [1/(x + y)] (1 + dy/dx) + (x dy/dx + y) = 0, using chain rule for ln and product rule for xy. The dy/dx terms are [1/(x + y)] dy/dx + x dy/dx, grouped, while - [1/(x + y)] - y is the other side. Solving for dy/dx gives - ([1/(x + y)] + y) / ([1/(x + y)] + x), as shown. Choice D is a tempting distractor if someone omits the x in the denominator, forgetting the dy/dx from xy. To spot implicit differentiation opportunities, identify logarithmic or product terms involving both variables.

Question 4

Given x2y+ln⁡y=5x^2y+\ln y=5x2y+lny=5, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. −2xyx2+1y-\dfrac{2xy}{x^2+\frac{1}{y}}−x2+y1​2xy​ (correct answer)
  2. −2xx2+1y-\dfrac{2x}{x^2+\frac{1}{y}}−x2+y1​2x​
  3. −2xyx2-\dfrac{2xy}{x^2}−x22xy​
  4. −2xy+1yx2-\dfrac{2xy+\frac{1}{y}}{x^2}−x22xy+y1​​
  5. −2xyx2−1y-\dfrac{2xy}{x^2-\frac{1}{y}}−x2−y1​2xy​

Explanation: This problem requires implicit differentiation to find dy/dx for the relation where y is not explicitly a function of x. Differentiating x^2 y + ln y = 5, x^2 y gives x^2 dy/dx + 2x y by product rule, ln y gives (1/y) dy/dx by chain rule. Dy/dx appears via product rule and chain rule on terms involving y. Group: 2x y + x^2 dy/dx + (1/y) dy/dx = 0, so (x^2 + 1/y) dy/dx = -2x y, dy/dx = -2x y / (x^2 + 1/y). A tempting distractor like choice C fails by omitting the 1/y in denominator and simplifying incorrectly. Recognize implicit differentiation when y appears in logs or products with x, making explicit solving tricky.

Question 5

If xy+yx=10x \sqrt{y} + y \sqrt{x} = 10xy​+yx​=10, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. -\dfrac{\sqrt{y} + \frac{y}{2 \sqrt{x}}}{\frac{x}{2 \sqrt{y}} + \sqrt{x}} (correct answer)
  2. -\dfrac{\sqrt{y} + \frac{y}{2 \sqrt{x}}}{\frac{x}{2 \sqrt{y}}}
  3. -\dfrac{\sqrt{y} + \frac{1}{2 \sqrt{x}}}{\frac{x}{2 \sqrt{y}} + \sqrt{x}}
  4. -\dfrac{\sqrt{y} + \frac{y}{2 \sqrt{x}}}{\frac{x}{2 \sqrt{y}} - \sqrt{x}}
  5. -\dfrac{\sqrt{y}}{\sqrt{x}}

Explanation: This problem requires implicit differentiation to find dy/dxdy/dxdy/dx for the relation where y is not explicitly a function of x. Differentiating xy+yx=10x \sqrt{y} + y \sqrt{x} = 10xy​+yx​=10, xyx \sqrt{y}xy​ gives y+x(1/(2y))dy/dx\sqrt{y} + x (1/(2 \sqrt{y})) dy/dxy​+x(1/(2y​))dy/dx by product and chain, yxy \sqrt{x}yx​ gives xdy/dx+y(1/(2x))\sqrt{x} dy/dx + y (1/(2 \sqrt{x}))x​dy/dx+y(1/(2x​)) by product and chain. Dy/dxDy/dxDy/dx terms come from product rules and chain rules on square roots. Group: y+(x/(2y))dy/dx+(y/(2x))+xdy/dx=0\sqrt{y} + (x/(2 \sqrt{y})) dy/dx + (y/(2 \sqrt{x})) + \sqrt{x} dy/dx = 0y​+(x/(2y​))dy/dx+(y/(2x​))+x​dy/dx=0, so [x/(2y)+x]dy/dx=−[y+y/(2x)][x/(2 \sqrt{y}) + \sqrt{x}] dy/dx = - [\sqrt{y} + y/(2 \sqrt{x})][x/(2y​)+x​]dy/dx=−[y​+y/(2x​)], matching A. A tempting distractor like choice B fails by omitting the x\sqrt{x}x​ in denominator, forgetting part of the grouping. Spot implicit differentiation needs when roots or other functions entwine x and y.

Question 6

For the relation x2+xy+y2=7x^2+xy+y^2=7x2+xy+y2=7, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. −2x+yx+2y-\dfrac{2x+y}{x+2y}−x+2y2x+y​ (correct answer)
  2. −2x+yx-\dfrac{2x+y}{x}−x2x+y​
  3. −2x+y2y-\dfrac{2x+y}{2y}−2y2x+y​
  4. −2xx+2y-\dfrac{2x}{x+2y}−x+2y2x​
  5. −2x+y+2yx-\dfrac{2x+y+2y}{x}−x2x+y+2y​

Explanation: This problem requires implicit differentiation to find dy/dx for the relation where y is not explicitly a function of x. Differentiating both sides with respect to x, the term x^2 gives 2x, xy gives x dy/dx + y via the product rule, and y^2 gives 2y dy/dx via the chain rule. These dy/dx terms appear because y is treated as a function of x, requiring the chain rule for derivatives involving y. Conceptually, group the terms with dy/dx together: (x + 2y) dy/dx = -(2x + y), then solve for dy/dx. A tempting distractor like choice B fails because it omits the 2y in the denominator, likely from forgetting the chain rule on y^2. To recognize when to use implicit differentiation, look for equations where y is intertwined with x and cannot be easily isolated.

Question 7

For the implicit curve ex+y+xy=4e^{x+y}+xy=4ex+y+xy=4, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. −ex+y+yex+y+x-\dfrac{e^{x+y}+y}{e^{x+y}+x}−ex+y+xex+y+y​ (correct answer)
  2. −ex+y+xex+y+y-\dfrac{e^{x+y}+x}{e^{x+y}+y}−ex+y+yex+y+x​
  3. −ex+y+yex+y-\dfrac{e^{x+y}+y}{e^{x+y}}−ex+yex+y+y​
  4. −ex+yex+y+x+y-\dfrac{e^{x+y}}{e^{x+y}+x+y}−ex+y+x+yex+y​
  5. −ex+y+yx-\dfrac{e^{x+y}+y}{x}−xex+y+y​

Explanation: This problem requires implicit differentiation to find dy/dx for the relation where y is not explicitly a function of x. Differentiating e^{x+y} + xy = 4, e^{x+y} gives e^{x+y} (1 + dy/dx) by chain rule, xy gives x dy/dx + y by product rule. Dy/dx terms arise from chain rule on the exponent and product rule on xy. Group: e^{x+y} + e^{x+y} dy/dx + y + x dy/dx = 0, so (e^{x+y} + x) dy/dx = - (e^{x+y} + y), dy/dx = - (e^{x+y} + y)/(e^{x+y} + x). A tempting distractor like choice C fails by omitting the +x in the denominator, possibly forgetting the dy/dx from xy. For transferable strategy, use implicit differentiation when the curve is defined implicitly and you need the slope without solving for y.

Question 8

Given y2sin⁡x+xcos⁡y=2y^2\sin x + x\cos y=2y2sinx+xcosy=2, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. dydx=−y2cos⁡x−cos⁡y2ysin⁡x−xsin⁡y\dfrac{dy}{dx}=\dfrac{-y^2\cos x-\cos y}{2y\sin x-x\sin y}dxdy​=2ysinx−xsiny−y2cosx−cosy​ (correct answer)
  2. dydx=−y2cos⁡x−cos⁡y2ysin⁡x+xsin⁡y\dfrac{dy}{dx}=\dfrac{-y^2\cos x-\cos y}{2y\sin x+x\sin y}dxdy​=2ysinx+xsiny−y2cosx−cosy​
  3. dydx=−y2cos⁡x+cos⁡y2ysin⁡x−xsin⁡y\dfrac{dy}{dx}=\dfrac{-y^2\cos x+\cos y}{2y\sin x-x\sin y}dxdy​=2ysinx−xsiny−y2cosx+cosy​
  4. dydx=−y2cos⁡x−cos⁡y2ysin⁡x\dfrac{dy}{dx}=\dfrac{-y^2\cos x-\cos y}{2y\sin x}dxdy​=2ysinx−y2cosx−cosy​
  5. dydx=−y2cos⁡x2ysin⁡x−xsin⁡y\dfrac{dy}{dx}=\dfrac{-y^2\cos x}{2y\sin x-x\sin y}dxdy​=2ysinx−xsiny−y2cosx​

Explanation: This problem involves implicit differentiation of y²sin x + x cos y = 2. For y²sin x, we use the product rule: y²·cos x + sin x·2y·dy/dx. For x cos y, we get x·(-sin y)·dy/dx + cos y·1. Setting up: y²cos x + 2y sin x·dy/dx - x sin y·dy/dx + cos y = 0. Collecting dy/dx terms: dy/dx(2y sin x - x sin y) = -y²cos x - cos y. Therefore, dy/dx = (-y²cos x - cos y)/(2y sin x - x sin y). Choice B has a plus sign in the denominator (2y sin x + x sin y), which is incorrect—the negative sign comes from differentiating cos y. Remember: when differentiating products involving both x and y, apply the product rule carefully and track the signs from derivatives of trigonometric functions.

Question 9

Given the implicit relation xy+y=3\dfrac{x}{y}+y=3yx​+y=3, what is dydx\dfrac{dy}{dx}dxdy​ in terms of xxx and yyy?

  1. −1y−xy-\dfrac{1}{y-\frac{x}{y}}−y−yx​1​ (correct answer)
  2. −1y+xy-\dfrac{1}{y+\frac{x}{y}}−y+yx​1​
  3. −yy−xy-\dfrac{y}{y-\frac{x}{y}}−y−yx​y​
  4. 1y−xy\dfrac{1}{y-\frac{x}{y}}y−yx​1​
  5. −11−xy2-\dfrac{1}{1-\frac{x}{y^2}}−1−y2x​1​

Explanation: This problem requires implicit differentiation of x/y + y = 3. Differentiating x/y using the quotient rule gives (y·1 - x·dy/dx)/y², and the full equation becomes (y - x·dy/dx)/y² + dy/dx = 0. Multiplying through by y² yields y - x·dy/dx + y²·dy/dx = 0, which rearranges to dy/dx(y² - x) = -y. Therefore, dy/dx = -y/(y² - x) = -1/(y - x/y). Choice B incorrectly has a plus sign in the denominator, which would result from a sign error when factoring. The recognition strategy is to clear fractions strategically and recognize when the result can be simplified by factoring out common terms.

Question 10

Given x+y=5\sqrt{x}+\sqrt{y}=5x​+y​=5, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?​

  1. dydx=−yx\dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{\sqrt{x}}dxdy​=−x​y​​ (correct answer)
  2. dydx=−xy\dfrac{dy}{dx}=-\dfrac{\sqrt{x}}{\sqrt{y}}dxdy​=−y​x​​
  3. dydx=yx\dfrac{dy}{dx}=\dfrac{\sqrt{y}}{\sqrt{x}}dxdy​=x​y​​
  4. dydx=−1x\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{x}}dxdy​=−x​1​
  5. dydx=−1y\dfrac{dy}{dx}=-\dfrac{1}{\sqrt{y}}dxdy​=−y​1​

Explanation: This problem involves implicit differentiation of √x + √y = 5. Rewriting as x^(1/2) + y^(1/2) = 5 and differentiating: (1/2)x^(-1/2) + (1/2)y^(-1/2)(dy/dx) = 0. This simplifies to 1/(2√x) + (1/(2√y))(dy/dx) = 0. Solving for dy/dx: (dy/dx)/(2√y) = -1/(2√x), which gives dy/dx = -(2√y)/(2√x) = -√y/√x. Choice B incorrectly inverts the ratio, placing √x in the numerator and √y in the denominator. When differentiating radical expressions, remember that d/dx[√y] = 1/(2√y)·(dy/dx), and the resulting fraction manipulation must preserve the correct order.

Question 11

If x3+y3=6xyx^3+y^3=6xyx3+y3=6xy, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?​

  1. 6y−3x23y2−6x\dfrac{6y-3x^2}{3y^2-6x}3y2−6x6y−3x2​ (correct answer)
  2. 6y−3x23y2\dfrac{6y-3x^2}{3y^2}3y26y−3x2​
  3. 6y−3x2−6x\dfrac{6y-3x^2}{-6x}−6x6y−3x2​
  4. 3x26y\dfrac{3x^2}{6y}6y3x2​
  5. 6y−3x23y2+6x\dfrac{6y-3x^2}{3y^2+6x}3y2+6x6y−3x2​

Explanation: This problem uses implicit differentiation on the equation x³ + y³ = 6xy. Differentiating both sides with respect to x gives 3x² + 3y²·dy/dx = 6x·dy/dx + 6y, where the product rule is applied to the right side. Rearranging to collect dy/dx terms yields 3y²·dy/dx - 6x·dy/dx = 6y - 3x², which factors as dy/dx(3y² - 6x) = 6y - 3x². Therefore, dy/dx = (6y - 3x²)/(3y² - 6x). Choice E incorrectly has a plus sign in the denominator, which would result from a sign error when moving terms. The strategy is to carefully track signs when collecting dy/dx terms to one side of the equation.

Question 12

A curve satisfies ey+xy=3e^{y}+xy=3ey+xy=3. Find dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y).​

  1. dydx=−yey+x\dfrac{dy}{dx}=-\dfrac{y}{e^{y}+x}dxdy​=−ey+xy​ (correct answer)
  2. dydx=−yey\dfrac{dy}{dx}=-\dfrac{y}{e^{y}}dxdy​=−eyy​
  3. dydx=−1ey+x\dfrac{dy}{dx}=-\dfrac{1}{e^{y}+x}dxdy​=−ey+x1​
  4. dydx=yey+x\dfrac{dy}{dx}=\dfrac{y}{e^{y}+x}dxdy​=ey+xy​
  5. dydx=−xyey+x\dfrac{dy}{dx}=-\dfrac{xy}{e^{y}+x}dxdy​=−ey+xxy​

Explanation: This problem requires implicit differentiation of e^y + xy = 3. When differentiating e^y, we apply the chain rule to get e^y(dy/dx), and when differentiating xy, we use the product rule to get y + x(dy/dx). The equation becomes e^y(dy/dx) + y + x(dy/dx) = 0, since the derivative of the constant 3 is zero. Factoring out dy/dx from the terms that contain it: dy/dx(e^y + x) = -y, which gives us dy/dx = -y/(e^y + x). Choice C incorrectly omits the y term that comes from the product rule when differentiating xy, treating it as if only x varies. When using implicit differentiation, always apply the product rule completely to terms like xy, remembering that both x and y can vary.

Question 13

For the relation x2y2+x=1x^2y^2+x=1x2y2+x=1, what is dydx\dfrac{dy}{dx}dxdy​ in terms of xxx and yyy?​

  1. dydx=−2xy2+12x2y\dfrac{dy}{dx}=-\dfrac{2xy^2+1}{2x^2y}dxdy​=−2x2y2xy2+1​ (correct answer)
  2. dydx=−2xy22x2y\dfrac{dy}{dx}=-\dfrac{2xy^2}{2x^2y}dxdy​=−2x2y2xy2​
  3. dydx=−2xy2+12x2\dfrac{dy}{dx}=-\dfrac{2xy^2+1}{2x^2}dxdy​=−2x22xy2+1​
  4. dydx=2xy2+12x2y\dfrac{dy}{dx}=\dfrac{2xy^2+1}{2x^2y}dxdy​=2x2y2xy2+1​
  5. dydx=−2x2y+12xy2\dfrac{dy}{dx}=-\dfrac{2x^2y+1}{2xy^2}dxdy​=−2xy22x2y+1​

Explanation: This problem requires implicit differentiation of x²y² + x = 1. For x²y², we apply the product rule treating it as (x²)(y²): 2x·y² + x²·2y(dy/dx), and differentiating x gives 1. The equation becomes 2xy² + 2x²y(dy/dx) + 1 = 0. Solving for dy/dx: 2x²y(dy/dx) = -2xy² - 1, which gives dy/dx = -(2xy² + 1)/(2x²y). Choice B incorrectly omits the 1 in the numerator that comes from differentiating the x term on the left side. When performing implicit differentiation, remember to differentiate every term in the equation, including standalone x terms that don't contain y.

Question 14

The relation xy+yx=10x\sqrt{y}+y\sqrt{x}=10xy​+yx​=10 defines yyy implicitly. What is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?

  1. dydx=−y+y2xx2y+x\dfrac{dy}{dx}=-\dfrac{\sqrt{y}+\frac{y}{2\sqrt{x}}}{\frac{x}{2\sqrt{y}}+\sqrt{x}}dxdy​=−2y​x​+x​y​+2x​y​​ (correct answer)
  2. dydx=−y+y2xx2y\dfrac{dy}{dx}=-\dfrac{\sqrt{y}+\frac{y}{2\sqrt{x}}}{\frac{x}{2\sqrt{y}}}dxdy​=−2y​x​y​+2x​y​​
  3. dydx=−yx\dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{\sqrt{x}}dxdy​=−x​y​​
  4. dydx=−x2y+xy+y2x\dfrac{dy}{dx}=-\dfrac{\frac{x}{2\sqrt{y}}+\sqrt{x}}{\sqrt{y}+\frac{y}{2\sqrt{x}}}dxdy​=−y​+2x​y​2y​x​+x​​
  5. dydx=−y+12xx2y+x\dfrac{dy}{dx}=-\dfrac{\sqrt{y}+\frac{1}{2\sqrt{x}}}{\frac{x}{2\sqrt{y}}+\sqrt{x}}dxdy​=−2y​x​+x​y​+2x​1​​

Explanation: This problem requires implicit differentiation to find dydx\dfrac{dy}{dx}dxdy​ for the relation xy+yx=10x \sqrt{y} + y \sqrt{x} = 10xy​+yx​=10. Differentiating both sides gives y+x(12y)dydx+xdydx+y(12x)=0\sqrt{y} + x \left( \frac{1}{2 \sqrt{y}} \right) \dfrac{dy}{dx} + \sqrt{x} \dfrac{dy}{dx} + y \left( \frac{1}{2 \sqrt{x}} \right) = 0y​+x(2y​1​)dxdy​+x​dxdy​+y(2x​1​)=0, using product and chain rules for each term. The dydx\dfrac{dy}{dx}dxdy​ terms are (x2y)dydx+xdydx\left( \frac{x}{2 \sqrt{y}} \right) \dfrac{dy}{dx} + \sqrt{x} \dfrac{dy}{dx}(2y​x​)dxdy​+x​dxdy​, grouped, while −y−y2x- \sqrt{y} - \frac{y}{2 \sqrt{x}}−y​−2x​y​ is on the other side. Solving yields −y+y2xx2y+x- \frac{ \sqrt{y} + \frac{y}{2 \sqrt{x}} }{ \frac{x}{2 \sqrt{y}} + \sqrt{x} }−2y​x​+x​y​+2x​y​​, matching the form. Choice B is tempting if someone forgets the xdydx\sqrt{x} \dfrac{dy}{dx}x​dxdy​ term, omitting it in the denominator. To recognize implicit differentiation, look for roots or other functions mixing xxx and yyy implicitly.

Question 15

If sin⁡(xy)+y=x2\sin(xy)+y=x^2sin(xy)+y=x2, what is dydx\dfrac{dy}{dx}dxdy​ in terms of xxx and yyy?

  1. dydx=2x−ycos⁡(xy)xcos⁡(xy)+1\dfrac{dy}{dx}=\dfrac{2x-y\cos(xy)}{x\cos(xy)+1}dxdy​=xcos(xy)+12x−ycos(xy)​ (correct answer)
  2. dydx=2x−ycos⁡(xy)xcos⁡(xy)\dfrac{dy}{dx}=\dfrac{2x-y\cos(xy)}{x\cos(xy)}dxdy​=xcos(xy)2x−ycos(xy)​
  3. dydx=2x−cos⁡(xy)xcos⁡(xy)+1\dfrac{dy}{dx}=\dfrac{2x-\cos(xy)}{x\cos(xy)+1}dxdy​=xcos(xy)+12x−cos(xy)​
  4. dydx=2x+ycos⁡(xy)xcos⁡(xy)+1\dfrac{dy}{dx}=\dfrac{2x+y\cos(xy)}{x\cos(xy)+1}dxdy​=xcos(xy)+12x+ycos(xy)​
  5. dydx=2x−ycos⁡(xy)xcos⁡(xy)+y\dfrac{dy}{dx}=\dfrac{2x-y\cos(xy)}{x\cos(xy)+y}dxdy​=xcos(xy)+y2x−ycos(xy)​

Explanation: This problem involves implicit differentiation of sin(xy) + y = x². When differentiating sin(xy), we use the chain rule to get cos(xy) times the derivative of xy, which by the product rule is y + x(dy/dx), giving us cos(xy)[y + x(dy/dx)]. The left side becomes y·cos(xy) + x·cos(xy)(dy/dx) + dy/dx, and the right side gives 2x. Rearranging to isolate dy/dx: x·cos(xy)(dy/dx) + dy/dx = 2x - y·cos(xy), which factors as dy/dx[x·cos(xy) + 1] = 2x - y·cos(xy). Therefore, dy/dx = (2x - y·cos(xy))/(x·cos(xy) + 1). Choice C incorrectly treats y as constant when differentiating sin(xy), missing the y term in the numerator. Remember that when differentiating composite functions like sin(xy), the chain rule requires differentiating the entire inner function xy.

Question 16

For the relation x2+xy+y2=9x^2+xy+y^2=9x2+xy+y2=9, what is dydx\dfrac{dy}{dx}dxdy​ at a general point (x,y)(x,y)(x,y)?​​

  1. dydx=−2x+yx+2y\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}dxdy​=−x+2y2x+y​ (correct answer)
  2. dydx=−2x+y1+2y\dfrac{dy}{dx}=-\dfrac{2x+y}{1+2y}dxdy​=−1+2y2x+y​
  3. dydx=−2xx+2y\dfrac{dy}{dx}=-\dfrac{2x}{x+2y}dxdy​=−x+2y2x​
  4. dydx=2x+yx+2y\dfrac{dy}{dx}=\dfrac{2x+y}{x+2y}dxdy​=x+2y2x+y​
  5. dydx=−2x+yx\dfrac{dy}{dx}=-\dfrac{2x+y}{x}dxdy​=−x2x+y​

Explanation: This problem asks for dy/dx from the implicit equation x² + xy + y² = 9, a conic section equation. Differentiating term by term: 2x + y + x(dy/dx) + 2y(dy/dx) = 0, where the xy term requires the product rule. Collecting dy/dx terms: x(dy/dx) + 2y(dy/dx) = -2x - y, which factors as dy/dx(x + 2y) = -(2x + y). Choice C incorrectly omits the y term in the numerator, likely from forgetting that d/dx(xy) = y + x(dy/dx), not just x(dy/dx). The recognition strategy is to always apply the product rule completely to mixed terms like xy, remembering that both factors contribute to the derivative.

Question 17

For the curve x+yln⁡x=6x+ y\ln x=6x+ylnx=6, what is dydx\dfrac{dy}{dx}dxdy​ in terms of xxx and yyy?

  1. dydx=−1+y/xln⁡x\dfrac{dy}{dx}=-\dfrac{1+y/x}{\ln x}dxdy​=−lnx1+y/x​ (correct answer)
  2. dydx=−1ln⁡x+y\dfrac{dy}{dx}=-\dfrac{1}{\ln x+y}dxdy​=−lnx+y1​
  3. dydx=−1+ln⁡xy\dfrac{dy}{dx}=-\dfrac{1+\ln x}{y}dxdy​=−y1+lnx​
  4. dydx=−1+yln⁡x\dfrac{dy}{dx}=-\dfrac{1+y}{\ln x}dxdy​=−lnx1+y​
  5. dydx=1+y/xln⁡x\dfrac{dy}{dx}=\dfrac{1+y/x}{\ln x}dxdy​=lnx1+y/x​

Explanation: This problem asks for dy/dx from x + y ln x = 6, where y is multiplied by ln x. Differentiating: 1 + (ln x)(dy/dx) + y·(1/x) = 0, using the product rule on y ln x and remembering that d/dx(ln x) = 1/x. Solving for dy/dx: (ln x)(dy/dx) = -1 - y/x, so dy/dx = -(1 + y/x)/(ln x). Choice D incorrectly writes the numerator as (1 + y) instead of (1 + y/x), missing that the derivative of ln x contributes a factor of 1/x to the y term. The strategy for logarithmic products is to carefully apply the product rule, remembering that d/dx(ln x) = 1/x, not just 1.

Question 18

For the implicit relation xsin⁡y+ycos⁡x=2x\sin y+y\cos x=2xsiny+ycosx=2, what is dydx\dfrac{dy}{dx}dxdy​ in terms of xxx and yyy?

  1. dydx=xsin⁡y−ysin⁡xxcos⁡y+cos⁡x\dfrac{dy}{dx}=\dfrac{x\sin y-y\sin x}{x\cos y+\cos x}dxdy​=xcosy+cosxxsiny−ysinx​
  2. dydx=ysin⁡x−sin⁡yxcos⁡y+cos⁡x\dfrac{dy}{dx}=\dfrac{y\sin x-\sin y}{x\cos y+\cos x}dxdy​=xcosy+cosxysinx−siny​
  3. dydx=−sin⁡y−ysin⁡xxcos⁡y+cos⁡x\dfrac{dy}{dx}=-\dfrac{\sin y-y\sin x}{x\cos y+\cos x}dxdy​=−xcosy+cosxsiny−ysinx​ (correct answer)
  4. dydx=−sin⁡y+ysin⁡xxcos⁡y+cos⁡x\dfrac{dy}{dx}=-\dfrac{\sin y+y\sin x}{x\cos y+\cos x}dxdy​=−xcosy+cosxsiny+ysinx​
  5. dydx=−sin⁡y−ysin⁡xxcos⁡y−cos⁡x\dfrac{dy}{dx}=-\dfrac{\sin y-y\sin x}{x\cos y-\cos x}dxdy​=−xcosy−cosxsiny−ysinx​

Explanation: This problem requires implicit differentiation of x sin y + y cos x = 2, involving products of x and y with trigonometric functions. Differentiating x sin y requires the product rule: sin y + x cos y(dy/dx), while differentiating y cos x gives cos x(dy/dx) - y sin x. Setting the total derivative to 0: sin y + x cos y(dy/dx) + cos x(dy/dx) - y sin x = 0, which rearranges to (x cos y + cos x)(dy/dx) = -sin y + y sin x. The common mistake in choice D is getting the wrong sign on y sin x, treating it as if it adds to sin y rather than subtracts. The key insight is that when differentiating y cos x, the derivative of cos x is -sin x, introducing a negative sign that makes the final term +y sin x in our rearrangement.

Question 19

For points (x,y)(x,y)(x,y) satisfying x2y+y3=7x^2y+y^3=7x2y+y3=7, what is dydx\dfrac{dy}{dx}dxdy​ in terms of xxx and yyy?

  1. dydx=−2xyx2+3y2\dfrac{dy}{dx}=-\dfrac{2xy}{x^2+3y^2}dxdy​=−x2+3y22xy​ (correct answer)
  2. dydx=−2xyx2\dfrac{dy}{dx}=-\dfrac{2xy}{x^2}dxdy​=−x22xy​
  3. dydx=−2xy1+3y2\dfrac{dy}{dx}=-\dfrac{2xy}{1+3y^2}dxdy​=−1+3y22xy​
  4. dydx=−2xx2+3y2\dfrac{dy}{dx}=-\dfrac{2x}{x^2+3y^2}dxdy​=−x2+3y22x​
  5. dydx=2xyx2+3y2\dfrac{dy}{dx}=\dfrac{2xy}{x^2+3y^2}dxdy​=x2+3y22xy​

Explanation: This problem requires implicit differentiation to find dy/dx from the equation x²y + y³ = 7. When we differentiate both sides with respect to x, the left side requires the product rule on x²y (giving 2xy + x²(dy/dx)) and the chain rule on y³ (giving 3y²(dy/dx)), while the right side gives 0. Collecting all terms with dy/dx on one side gives us x²(dy/dx) + 3y²(dy/dx) = -2xy, which factors as (x² + 3y²)(dy/dx) = -2xy. A common error is forgetting the product rule on x²y and only getting x²(dy/dx), which would lead to the incorrect answer B. The key strategy is to remember that every y term generates a dy/dx factor when differentiated, and to carefully apply the product rule to mixed terms like x²y.

Question 20

Given x3+y3=3xyx^3+y^3=3xyx3+y3=3xy, find dydx\dfrac{dy}{dx}dxdy​ as a function of xxx and yyy.

  1. dydx=3x2−3y3x−3y2\dfrac{dy}{dx}=\dfrac{3x^2-3y}{3x-3y^2}dxdy​=3x−3y23x2−3y​ (correct answer)
  2. dydx=3x2−3y3y2−3x\dfrac{dy}{dx}=\dfrac{3x^2-3y}{3y^2-3x}dxdy​=3y2−3x3x2−3y​
  3. dydx=3x2−3y3x+3y2\dfrac{dy}{dx}=\dfrac{3x^2-3y}{3x+3y^2}dxdy​=3x+3y23x2−3y​
  4. dydx=3x23y2\dfrac{dy}{dx}=\dfrac{3x^2}{3y^2}dxdy​=3y23x2​
  5. dydx=3x2−3y3x−3y\dfrac{dy}{dx}=\dfrac{3x^2-3y}{3x-3y}dxdy​=3x−3y3x2−3y​

Explanation: This problem asks us to find dy/dx for the implicit relation x³ + y³ = 3xy, known as the folium of Descartes. Differentiating both sides: 3x² + 3y²(dy/dx) = 3y + 3x(dy/dx), where the right side uses the product rule on 3xy. Rearranging to collect dy/dx terms: 3y²(dy/dx) - 3x(dy/dx) = 3y - 3x², which factors as dy/dx(3y² - 3x) = 3y - 3x². The incorrect answer B reverses the signs in the denominator, getting 3y² - 3x instead of 3x - 3y² after moving terms. Remember that when moving a dy/dx term from right to left, its coefficient changes sign, so 3x(dy/dx) on the right becomes -3x(dy/dx) on the left, leading to the correct denominator 3x - 3y².