AP Calculus BC Quiz: Exploring Behaviors Of Implicit Relations

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 BC.

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.

Question 1

1. $\dfrac{-\cos y-y\cos x}{-x\sin y+\sin x}$ (correct answer)
2. $\dfrac{-\cos y-y\cos x}{x\sin y+\sin x}$
3. $\dfrac{-\cos y+y\cos x}{-x\sin y+\sin x}$
4. $\dfrac{-\cos y-y\cos x}{-x\sin y+\sin x\,\dfrac{dy}{dx}}$
5. $\dfrac{-x\sin y+\sin x}{-\cos y-y\cos x}$

Explanation: This problem requires implicit differentiation to find dy/dx for the relation x cos y + y sin x = 3. Chain and product rules introduce dy/dx for trigonometric terms with y. Differentiation yields cos y - x sin y dy/dx + sin x dy/dx + y cos x = 0. Terms are grouped by isolating dy/dx factors, giving (-cos y - y cos x)/(-x sin y + sin x). Choice B fails as a distractor with an incorrect sign in the denominator, altering the expression. Recognize this technique for implicit trig relations where x and y are arguments or coefficients in sine and cosine.

Question 2

1. $\dfrac{-2x-y}{x+\cos y}$ (correct answer)
2. $\dfrac{-2x+y}{x+\cos y}$
3. $\dfrac{-2x-y}{x-\cos y}$
4. $\dfrac{-2x}{x+\cos y}$
5. $\dfrac{-2x-y\,\dfrac{dy}{dx}}{x+\cos y}$

Explanation: This problem requires implicit differentiation to find dy/dx for the implicitly defined relation x² + x y + sin y = 0. When differentiating, terms like sin y produce cos y dy/dx, and x y yield x dy/dx + y. These dy/dx terms appear because y is a function of x, applying chain and product rules. To solve, group dy/dx terms (x dy/dx + cos y dy/dx) and constants (-2x - y), then isolate dy/dx. A tempting distractor like choice E fails because it leaves dy/dx unsolved in the numerator. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.

Question 3

1. $\dfrac{-2xy}{x^2-\cos y}$ (correct answer)
2. $\dfrac{-2xy}{x^2+\cos y}$
3. $\dfrac{2xy}{x^2-\cos y}$
4. $\dfrac{-2x}{x^2-\cos y}$
5. $\dfrac{-2xy-\cos y\,\dfrac{dy}{dx}}{x^2}$

Explanation: This problem requires implicit differentiation to find dy/dx for the relation x² y = sin y. Differentiating: 2 x y + x² dy/dx = cos y dy/dx. Dy/dx from product and trig chain. Grouping: 2 x y = dy/dx (cos y - x²), dy/dx = 2 x y / (cos y - x²) = - 2 x y / (x² - cos y), choice A. Choice B has + cos y, denominator error. Spot in polynomial-trig relations.

Question 4

1. $\dfrac{-2xe^y}{x^2e^y+1}$ (correct answer)
2. $\dfrac{2xe^y}{x^2e^y+1}$
3. $\dfrac{-2xe^y}{x^2e^y-1}$
4. $\dfrac{-2x}{x^2e^y+1}$
5. $\dfrac{-2xe^y-y\,\dfrac{dy}{dx}}{x^2e^y+1}$

Explanation: This problem requires implicit differentiation to find dy/dx for the relation x² e^y + y = 4. Differentiating both sides with respect to x introduces dy/dx via the chain rule for terms involving y. The term x² e^y requires the product rule, yielding 2x e^y + x² e^y dy/dx, while the standalone y differentiates to dy/dx. Collecting like terms groups the dy/dx factors together as (x² e^y + 1) dy/dx = -2x e^y, allowing us to solve for dy/dx = -2x e^y / (x² e^y + 1). A tempting distractor like choice D forgets the dy/dx from the y term, incorrectly omitting the +1 in the denominator. To recognize implicit differentiation opportunities, look for equations defining y implicitly without solving for y explicitly.

Question 5

1. $\dfrac{-2xy-y\sec^2(xy)}{x^2+x\sec^2(xy)}$ (correct answer)
2. $\dfrac{2xy+y\sec^2(xy)}{x^2+x\sec^2(xy)}$
3. $\dfrac{-2xy-y\sec^2(xy)}{x^2-x\sec^2(xy)}$
4. $\dfrac{-2xy}{x^2+x\sec^2(xy)}$
5. $\dfrac{-2xy-y\sec^2(xy)\,\dfrac{dy}{dx}}{x^2+x\sec^2(xy)}$

Explanation: This problem requires implicit differentiation to find dy/dx for the relation x² y + tan(x y) = 0. Differentiating: 2 x y + x² dy/dx + sec²(x y) (y + x dy/dx) = 0. Dy/dx from product and chain rules. Grouping: 2 x y + y sec²(x y) + dy/dx (x² + x sec²(x y)) = 0, dy/dx = - (2 x y + y sec²) / (x² + x sec²) = - y (2 x + sec²) / [x (x + sec²)], but simplified to choice A. Choice B has positive numerator, from sign error. Spot in trig functions of products.

Question 6

1. $\dfrac{\frac{1}{x^2 y} - 2x}{2y + \frac{1}{x y^2}}$
2. $\dfrac{-\frac{1}{x^2 y} - 2x}{2y - \frac{1}{x y^2}}$
3. $\dfrac{-\frac{1}{x^2 y} - 2x}{2y + \frac{1}{x y^2}}$ (correct answer)
4. $\dfrac{-2x}{2y + \frac{1}{x y^2}}$
5. $\dfrac{-2x + \frac{1}{x^2 y} \dfrac{dy}{dx}}{2y + \frac{1}{x y^2}}$

Explanation: This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the relation $x^2 + y^2 = \dfrac{1}{xy}$. Differentiating: $2x + 2y \dfrac{dy}{dx} = -\dfrac{1}{(xy)^2} (y + x \dfrac{dy}{dx})$. Expand: right = $- \dfrac{y + x \dfrac{dy}{dx}}{x^2 y^2}$. So $2x + 2y \dfrac{dy}{dx} + \dfrac{y + x \dfrac{dy}{dx}}{x^2 y^2} = 0$. To combine, multiply through by $x^2 y^2$: $2x \cdot x^2 y^2 + 2y \dfrac{dy}{dx} x^2 y^2 + y + x \dfrac{dy}{dx} = 0$? Wait, better: the equation is $2x + 2y y' = - \dfrac{1}{x^2 y^2} (y + x y')$. Let's group: $2x + 2y y' + \dfrac{y}{x^2 y^2} + \dfrac{x y'}{x^2 y^2} = 0$. Simplify: $2x + \dfrac{1}{x^2 y} + y' (2y + \dfrac{1}{x y^2}) = 0$. Wait, $\dfrac{y}{x^2 y^2} = \dfrac{1}{x^2 y}$, $\dfrac{x}{x^2 y^2} = \dfrac{1}{x y^2}$. Yes, so $\dfrac{dy}{dx} = - \dfrac{2x + \dfrac{1}{x^2 y}}{2y + \dfrac{1}{x y^2}}$, which is choice C. Choice A has wrong signs. Recognize in reciprocal relations.

Question 7

1. $\dfrac{-2x}{\sec^2 y-3}$ (correct answer)
2. $\dfrac{-2x}{\sec^2 y+3}$
3. $\dfrac{2x}{\sec^2 y-3}$
4. $\dfrac{-2x-\sec^2 y\,\dfrac{dy}{dx}}{-3}$
5. $\dfrac{-2}{\sec^2 y-3}$

Explanation: This problem requires implicit differentiation to find dy/dx for the implicitly defined relation x² + tan y = 3y. When differentiating, terms like tan y produce sec² y dy/dx via the chain rule, and 3y yields 3 dy/dx. These dy/dx terms appear because y is a function of x, necessitating the chain rule. To solve, group dy/dx terms (sec² y dy/dx - 3 dy/dx) and constants (-2x), then isolate dy/dx. A tempting distractor like choice D fails because it leaves dy/dx unsolved in the numerator. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.

Question 8

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

Explanation: This problem requires implicit differentiation of √x + √y = 5 to find dy/dx. Rewriting as x^(1/2) + y^(1/2) = 5 and differentiating gives (1/2)x^(-1/2) + (1/2)y^(-1/2)(dy/dx) = 0. Multiplying through by 2 to clear fractions: x^(-1/2) + y^(-1/2)(dy/dx) = 0, so y^(-1/2)(dy/dx) = -x^(-1/2). Therefore, dy/dx = -x^(-1/2)/y^(-1/2) = -y^(1/2)/x^(1/2) = -√y/√x. A common error is to invert the fraction incorrectly, getting -√x/√y instead. The key insight is that when dividing by y^(-1/2), you multiply by y^(1/2), and the pattern x^(-1/2)/y^(-1/2) = y^(1/2)/x^(1/2) follows from the rule for dividing powers.

Question 9

1. $-\dfrac{\cos y+y\cos x}{-x\sin y+\sin x}$ (correct answer)
2. $-\dfrac{\cos y+y\cos x}{x\sin y+\sin x}$
3. $-\dfrac{-x\sin y+\sin x}{\cos y+y\cos x}$
4. $\dfrac{\cos y+y\cos x}{-x\sin y+\sin x}$
5. $-\dfrac{\cos y+\cos x}{-x\sin y+\sin x}$

Explanation: This problem uses implicit differentiation on x·cos(y) + y·sin(x) = 1 to find dy/dx. Differentiating requires the product rule on both terms: d/dx[x·cos(y)] = cos(y) + x·(-sin(y))·(dy/dx) and d/dx[y·sin(x)] = (dy/dx)·sin(x) + y·cos(x). Setting the sum equal to 0: cos(y) - x·sin(y)·(dy/dx) + (dy/dx)·sin(x) + y·cos(x) = 0. Collecting dy/dx terms: (sin(x) - x·sin(y))·(dy/dx) = -(cos(y) + y·cos(x)), so dy/dx = -(cos(y) + y·cos(x))/(sin(x) - x·sin(y)). A common error is forgetting the negative sign when cos(y) appears in the derivative of cos(y), leading to sign errors. The recognition strategy is to carefully track signs through trigonometric derivatives and note that the denominator can be written as -(x·sin(y) - sin(x)) = -x·sin(y) + sin(x).

Question 10

1. $\dfrac{y-2x}{x-\sec^2 y}$
2. $\dfrac{2x-y}{x-\sec^2 y}$ (correct answer)
3. $\dfrac{2x}{x}$
4. $\dfrac{2x-y}{x+\sec^2 y}$
5. $\dfrac{2x- y - x\,\dfrac{dy}{dx}}{\sec^2 y}$

Explanation: This problem involves implicit differentiation of x² + tan y = xy. Differentiating both sides: 2x + sec²y·dy/dx = x·dy/dx + y (using chain rule for tan y and product rule for xy). Rearranging to collect dy/dx terms: sec²y·dy/dx - x·dy/dx = y - 2x, which factors as dy/dx(sec²y - x) = y - 2x. Therefore, dy/dx = (y - 2x)/(sec²y - x), but multiplying numerator and denominator by -1 gives (2x - y)/(x - sec²y), which is choice B. Choice A has the signs reversed incorrectly. The recognition strategy is to remember that d/dx[tan y] = sec²y·dy/dx and to carefully track signs when rearranging terms.

Question 11

1. $\dfrac{-e^{x+y}-2xy}{e^{x+y}+x^2}$ (correct answer)
2. $\dfrac{-e^{x+y}-2x^2y}{e^{x+y}+x^2}$
3. $\dfrac{-e^{x+y}-2xy}{e^{x+y}}$
4. $\dfrac{-e^{x+y}-2xy}{e^{x+y}+x^2\,\dfrac{dy}{dx}}$
5. $\dfrac{-e^{x+y}-2x}{e^{x+y}+x^2}$

Explanation: This problem requires implicit differentiation to find dy/dx for the relation e^{x+y} + x²y = 5. Dy/dx emerges from the chain rule applied to exponential and product terms involving y. The derivative of e^{x+y} is e^{x+y}(1 + dy/dx), and x²y gives 2xy + x² dy/dx, summing to zero. We conceptually group dy/dx coefficients together, solving as (-e^{x+y} - 2xy)/(e^{x+y} + x²). Choice C is a tempting distractor but omits the x² in the denominator, likely from forgetting the product rule's second part. Recognize implicit differentiation when exponential or polynomial mixtures of x and y prevent easy isolation of y.

Question 12

1. $\dfrac{-1}{1+4y\sqrt{x+y}}$ (correct answer)
2. $\dfrac{-1}{2\sqrt{x+y}+2y}$
3. $\dfrac{-1}{1+4y\sqrt{x+y}\,\dfrac{dy}{dx}}$
4. $\dfrac{-1}{1-4y\sqrt{x+y}}$
5. $\dfrac{1}{1+4y\sqrt{x+y}}$

Explanation: This problem requires implicit differentiation of √(x + y) + y² = 5. The square root term uses the chain rule: d/dx[√(x + y)] = 1/(2√(x + y))·(1 + dy/dx). The complete differentiation yields: (1/(2√(x + y)))(1 + dy/dx) + 2y(dy/dx) = 0. Multiplying through by 2√(x + y) to clear the fraction: 1 + dy/dx + 4y√(x + y)(dy/dx) = 0. Factoring out dy/dx: 1 + dy/dx[1 + 4y√(x + y)] = 0, which gives dy/dx = -1/(1 + 4y√(x + y)). Choice C incorrectly places dy/dx in the denominator alongside the other terms. The key insight is to clear fractions before collecting dy/dx terms to avoid algebraic errors.

Question 13

1. $\dfrac{-2x-y}{x+2y}$ (correct answer)
2. $\dfrac{-2x-y}{1+2y}$
3. $\dfrac{-2x-y}{x+2y\,\dfrac{dy}{dx}}$
4. $\dfrac{2x+y}{x+2y}$
5. $\dfrac{-2x+y}{x+2y}$

Explanation: This problem requires implicit differentiation of x² + xy + y² = 7. Differentiating term by term: 2x + (y + x(dy/dx)) + 2y(dy/dx) = 0, where the middle term uses the product rule. Collecting dy/dx terms gives x(dy/dx) + 2y(dy/dx) = -2x - y, which factors as (x + 2y)(dy/dx) = -2x - y. Therefore, dy/dx = (-2x - y)/(x + 2y). Choice C shows a common algebraic error of leaving dy/dx in the denominator instead of factoring it out. The key recognition pattern is that every term containing y will contribute a dy/dx term when differentiated.

Question 14

1. $\dfrac{1-y\cos(xy)}{x\cos(xy)+1}$ (correct answer)
2. $\dfrac{1-y\cos(xy)}{x\cos(xy)}$
3. $\dfrac{1+y\cos(xy)}{x\cos(xy)+1}$
4. $\dfrac{1-y\cos(xy)}{x\cos(xy)+\dfrac{dy}{dx}}$
5. $\dfrac{y\cos(xy)-1}{x\cos(xy)+1}$

Explanation: This problem requires implicit differentiation of sin(xy) + y = x. When differentiating sin(xy), we must use the chain rule combined with the product rule, giving cos(xy)·(y + x(dy/dx)). The full differentiation yields cos(xy)·y + cos(xy)·x(dy/dx) + dy/dx = 1. Collecting dy/dx terms gives cos(xy)·x + 1 = 1 - y·cos(xy). Therefore, dy/dx = (1 - y·cos(xy))/(x·cos(xy) + 1). Choice D incorrectly places dy/dx in the denominator rather than factoring it out. The key insight is recognizing that differentiating composite functions like sin(xy) requires both chain and product rules.

Question 15

1. $\dfrac{y^2}{x^2}$
2. $-\dfrac{y^2}{x^2}$ (correct answer)
3. $-\dfrac{x^2}{y^2}$
4. $\dfrac{1}{x^2}$
5. $-\dfrac{\frac{1}{x^2}}{\frac{1}{y^2}\,\dfrac{dy}{dx}}$

Explanation: This problem requires implicit differentiation to find dy/dx for the relation 1/x + 1/y = 1. Differentiating gives -1/x² - (1/y²) dy/dx = 0. The dy/dx term appears from the chain rule on 1/y, treated as y^{-1}. Solving isolates dy/dx = - (1/x²) / (-1/y²) = - y² / x². Choice A omits the negative sign, possibly from forgetting the chain rule's impact on signs. Identify implicit differentiation when reciprocals or fractions involve both variables.

Question 16

1. $\dfrac{-y\cos(xy)}{x\cos(xy)+1}$ (correct answer)
2. $\dfrac{-y\cos(xy)}{x\cos(xy)}$
3. $\dfrac{y\cos(xy)}{x\cos(xy)+1}$
4. $\dfrac{-\cos(xy)}{x\cos(xy)+1}$
5. $\dfrac{-y\cos(xy)-\dfrac{dy}{dx}}{x\cos(xy)+1}$

Explanation: This problem requires implicit differentiation to find dy/dx for the implicitly defined relation sin(xy) + y = 4. When differentiating, composite terms like sin(xy) produce cos(xy)(x dy/dx + y) via the chain and product rules, and y yields dy/dx. These dy/dx terms appear because y is a function of x, requiring the chain rule for nested functions. To solve, group dy/dx terms (x cos(xy) dy/dx + dy/dx) and constants (-y cos(xy)), then isolate dy/dx. A tempting distractor like choice E fails because it leaves dy/dx unsolved in the expression. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.

Question 17

1. $\dfrac{-2x-y}{x+2y}$ (correct answer)
2. $\dfrac{-2x-y\,\dfrac{dy}{dx}}{x+2y}$
3. $\dfrac{-2x-y}{2y}$
4. $\dfrac{-2x+y}{x+2y}$
5. $\dfrac{-2x-y}{x}$

Explanation: This problem requires implicit differentiation to find dy/dx for the implicitly defined relation x² + xy + y² = 7. When differentiating, terms like y² produce 2y dy/dx via the chain rule, and xy requires the product rule, yielding y + x dy/dx. These dy/dx terms appear because y is treated as a function of x, necessitating the chain rule for y-dependent terms. To solve, collect all dy/dx terms on one side (x dy/dx + 2y dy/dx) and the constant terms on the other (-2x - y), then isolate dy/dx by dividing. A tempting distractor like choice B fails because it incorrectly leaves dy/dx in the numerator without solving for it. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.

Question 18

1. $\dfrac{1- y\cos(xy)}{x\cos(xy)-1}$
2. $\dfrac{y\cos(xy)-1}{x\cos(xy)-1}$ (correct answer)
3. $\dfrac{\cos(xy)}{1}$
4. $\dfrac{y\cos(xy)-1}{x\cos(xy)}$
5. $\dfrac{\cos(xy)(y+x\,\dfrac{dy}{dx})-1}{1}$

Explanation: This problem requires implicit differentiation of sin(xy) = x + y. Using the chain rule on the left side: d/dx[sin(xy)] = cos(xy)·d/dx[xy] = cos(xy)·(x·dy/dx + y·1) = x·cos(xy)·dy/dx + y·cos(xy). The right side simply gives 1 + dy/dx. Setting them equal: x·cos(xy)·dy/dx + y·cos(xy) = 1 + dy/dx. Rearranging to collect dy/dx terms: x·cos(xy)·dy/dx - dy/dx = 1 - y·cos(xy), which factors as dy/dx(x·cos(xy) - 1) = 1 - y·cos(xy). Therefore, dy/dx = (1 - y·cos(xy))/(x·cos(xy) - 1), but multiplying numerator and denominator by -1 gives (y·cos(xy) - 1)/(x·cos(xy) - 1), which is choice B. Choice A incorrectly flips only the numerator's sign. The key is remembering that chain rule on trig functions produces the derivative of the inner function as a factor.

Question 19

1. $-\dfrac{ye^{x}+e^{y}}{e^{x}+xe^{y}}$ (correct answer)
2. $-\dfrac{ye^{x}+e^{y}}{e^{x}+xe^{y}\,\dfrac{dy}{dx}}$
3. $-\dfrac{e^{x}+xe^{y}}{ye^{x}+e^{y}}$
4. $-\dfrac{ye^{x}+xe^{y}}{e^{x}+e^{y}}$
5. $-\dfrac{ye^{x}}{e^{x}+xe^{y}}$

Explanation: This problem requires implicit differentiation of y·e^x + x·e^y = 10 to find dy/dx. Using the product rule on both terms: d/dx[y·e^x] = (dy/dx)·e^x + y·e^x and d/dx[x·e^y] = e^y + x·e^y·(dy/dx). Setting the sum equal to 0: (dy/dx)·e^x + y·e^x + e^y + x·e^y·(dy/dx) = 0. Collecting dy/dx terms: e^x·(dy/dx) + x·e^y·(dy/dx) = -(y·e^x + e^y), which factors as (e^x + x·e^y)(dy/dx) = -(y·e^x + e^y). Therefore, dy/dx = -(y·e^x + e^y)/(e^x + x·e^y). A common mistake is to include dy/dx in the denominator of the final answer (as in choice B), forgetting that we've already solved for it. The key insight is that once you've isolated dy/dx algebraically, it shouldn't appear anywhere in your final expression.

Question 20

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

Explanation: This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the implicitly defined relation $x + y + \sqrt{xy} = 9$. When differentiating, terms like $\sqrt{xy}$ produce $\frac{1}{2\sqrt{xy}} (y + x \dfrac{dy}{dx})$ via chain and product rules. These $\dfrac{dy}{dx}$ terms appear because y is a function of x, requiring combined rules for composite terms. To solve, group $\dfrac{dy}{dx}$ terms ($\dfrac{dy}{dx} + \frac{x}{2\sqrt{xy}} \dfrac{dy}{dx}$) and constants ($-1 - \frac{y}{2\sqrt{xy}}$), then isolate $\dfrac{dy}{dx}$. A tempting distractor like choice E fails because it leaves $\dfrac{dy}{dx}$ unsolved in the expression. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.