AP Calculus AB Quiz: Implicit Differentiation

What this quiz covers

This quiz focuses on Implicit Differentiation, 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.

Question 1

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

Explanation: This problem uses implicit differentiation for the level curve e^x + e^y = xy to find dy/dx. Differentiating yields e^x + e^y dy/dx = y + x dy/dx, with chain rule on e^y and product rule on xy. Dy/dx terms are e^y dy/dx - x dy/dx, grouping to (e^y - x) dy/dx, and constants y - e^x. Thus, dy/dx = (y - e^x)/(e^y - x), noting the negative in numerator for solving. Choice C is tempting but fails by using +x in denominator, likely forgetting to move x dy/dx correctly. Spot implicit differentiation in equations with exponentials of x and y equated to products.

Question 2

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

Explanation: This problem involves implicit differentiation for x ln y + y ln x = 10 to find dy/dx. Differentiating gives ln y + x (1/y) dy/dx + ln x dy/dx + y (1/x) = 0, with product rules. Dy/dx terms are (x/y) dy/dx + ln x dy/dx, factoring to (x/y + ln x) dy/dx. Constants are ln y + y/x, so dy/dx = - (ln y + y/x) / (x/y + ln x). Choice B tempts with x/y in numerator but fails by mismatching terms from product rule. Recognize this in symmetric logarithmic terms like x ln y and y ln x.

Question 3

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

Explanation: This problem requires implicit differentiation to find the derivative dy/dx for the implicitly defined curve x² + xy + y² = 7. Differentiating both sides with respect to x yields 2x + (x dy/dx + y) + 2y dy/dx = 0, where the product rule is applied to the xy term. The terms involving dy/dx are x dy/dx + 2y dy/dx, which factor to (x + 2y) dy/dx, while the remaining terms are 2x + y. Solving for dy/dx gives dy/dx = -(2x + y)/(x + 2y), grouping the dy/dx terms on one side and constants on the other. A tempting distractor like choice D fails because it incorrectly simplifies the denominator to x + y instead of x + 2y, likely from mishandling the derivative of y². To recognize when to use implicit differentiation, look for equations where y is not isolated as a function of x but both variables are mixed together.

Question 4

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

Explanation: This problem requires implicit differentiation for $\sqrt{x^2 + y^2} = x + y$ to find $\frac{dy}{dx}$. Differentiating gives $\frac{1}{2\sqrt{x^2 + y^2}} (2x + 2y \frac{dy}{dx}) = 1 + \frac{dy}{dx}$, chain rule on left. Simplifies to $\frac{x + y \frac{dy}{dx}}{\sqrt{x^2 + y^2}} = 1 + \frac{dy}{dx}$. Dy/dx terms: $[ \frac{y}{\sqrt{x^2 + y^2}} - 1 ] \frac{dy}{dx} = 1 - \frac{x}{\sqrt{x^2 + y^2}}$, but A has numerator $\frac{x}{\sqrt{x^2 + y^2}} - 1$ over $1 - \frac{y}{\sqrt{x^2 + y^2}}$. Yes, matching after rearrangement. Choice B tempts with + in denominator but fails by not accounting for the sign when isolating dy/dx. Recognize in distance-like equations with square roots of sums.

Question 5

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

Explanation: This problem requires implicit differentiation for ln(x + y) = xy to find dy/dx. Differentiating gives [1/(x + y)] (1 + dy/dx) = y + x dy/dx, applying chain rule to ln and product to xy. Dy/dx terms are [1/(x + y)] dy/dx - x dy/dx, grouping to {1/(x + y) - x} dy/dx. Constants are y - 1/(x + y), so dy/dx = [y (x + y) - 1]/[1 - x (x + y)] after multiplying numerator and denominator by x + y. Choice B is tempting but fails with a sign error in the numerator, using 1 - y(x + y) instead. Identify implicit differentiation in logarithmic equations involving sums like x + y.

Question 6

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

Explanation: This problem involves implicit differentiation for y² ln x + x ln y = 3 to find dy/dx. Differentiating gives 2y dy/dx ln x + y² (1/x) + ln y + x (1/y) dy/dx = 0, applying product and chain rules. Dy/dx terms are 2y ln x dy/dx + (x/y) dy/dx, factoring to (2y ln x + x/y) dy/dx. Constants are y²/x + ln y, so dy/dx = - (y²/x + ln y) / (2y ln x + x/y). Choice B is tempting but fails by omitting y in the denominator's first term, likely a chain rule error. Recognize this in logarithmic products with both x and y.

Question 7

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

Explanation: This problem requires implicit differentiation for $y \sqrt{x} + x \sqrt{y} = 6$ to find $\dfrac{dy}{dx}$. Differentiating gives $y \left( \frac{1}{2 \sqrt{x}} \right) + \sqrt{x} \dfrac{dy}{dx} + \sqrt{y} + x \left( \frac{1}{2 \sqrt{y}} \right) \dfrac{dy}{dx} = 0$, product and chain. $\dfrac{dy}{dx}$ terms: $\sqrt{x} \dfrac{dy}{dx} + \left( \frac{x}{2 \sqrt{y}} \right) \dfrac{dy}{dx}$, grouping to $\left( \sqrt{x} + \frac{x}{2 \sqrt{y}} \right) \dfrac{dy}{dx}$. Constants: $\frac{y}{2 \sqrt{x}} + \sqrt{y}$, so $\dfrac{dy}{dx} = - \left( \frac{y}{2 \sqrt{x}} + \sqrt{y} \right) / \left( \sqrt{x} + \frac{x}{2 \sqrt{y}} \right)$. Choice B tempts with $-$ in denominator but fails by changing sign incorrectly. Recognize in products of variables and their roots.

Question 8

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

Explanation: This problem involves implicit differentiation for x² y + x y² = 8 to find dy/dx. Differentiating gives 2x y + x² dy/dx + y² + 2 x y dy/dx = 0, product rules on both terms. Dy/dx terms: x² dy/dx + 2 x y dy/dx, factoring to (x² + 2 x y) dy/dx. Constants: 2x y + y², so dy/dx = - (2x y + y²) / (x² + 2 x y). Choice B tempts with x² + x y in denominator but fails by undercounting the coefficient from derivative of y². Recognize in factored forms like xy (x + y) = 8.

Question 9

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

Explanation: This problem uses implicit differentiation for x^4 + y^4 = 2 x² y² + 3 to find dy/dx. Differentiating gives 4 x³ + 4 y³ dy/dx = 4 x y² + 2 x² (2 y dy/dx), product on right. Dy/dx terms: 4 y³ dy/dx - 4 x² y dy/dx, grouping to (4 y³ - 4 x² y) dy/dx. Constants: 4 x³ - 4 x y², so dy/dx = (4 x y² - 4 x³) / (4 y³ - 4 x² y) = - (x (x² - y²)) / (y (y² - x²)), matching D. Choice A tempts without negative but fails to account for sign when solving. Recognize in high-power symmetric equations.

Question 10

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

Explanation: This problem uses implicit differentiation for x²y + tan y = 5 to find dy/dx. Differentiating yields 2x y + x² dy/dx + sec² y dy/dx = 0, with product rule on x²y and chain on tan y. Dy/dx terms are x² dy/dx + sec² y dy/dx, factoring to (x² + sec² y) dy/dx. The constant term is 2x y, so dy/dx = -2x y / (x² + sec² y). Choice E tempts by omitting the negative sign but fails as differentiation requires moving positive terms to negative. Recognize this in equations mixing polynomials and trig functions of y.

Question 11

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

Explanation: This problem requires implicit differentiation for √x + √y = √(xy) to find dy/dx. Differentiating gives (1/(2√x)) + (1/(2√y)) dy/dx = (1/(2√(xy))) (y + x dy/dx), using chain rule on square roots. Dy/dx terms are (1/(2√y)) dy/dx - (1/(2√(xy))) x dy/dx, but solving yields the form in D after simplification. Grouping shows numerator (√y/(2x) - 1/(2√x)) and denominator (1/(2√y) - √x/(2y)). Choice C tempts with similar terms but fails by using incorrect coefficients like √y/(2x) wrongly. Recognize implicit differentiation in equations with roots of products like √(xy).

Question 12

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

Explanation: This problem involves implicit differentiation for x² y² + x + y = 0 to find dy/dx. Differentiating gives 2x y² + x² (2 y dy/dx) + 1 + dy/dx = 0, product on x² y². Dy/dx terms: 2 x² y dy/dx + dy/dx, factoring to (2 x² y + 1) dy/dx. Constants: 2x y² + 1, so dy/dx = - (2x y² + 1) / (2 x² y). Choice C tempts by simplifying denominator to 2xy but fails to include the +1 from derivative of y. Recognize in equations with high-degree products like x² y².

Question 13

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

Explanation: This problem uses implicit differentiation for x² + x y = ln y to find dy/dx. Differentiating gives 2x + y + x dy/dx = (1/y) dy/dx, product on xy, chain on ln y. Dy/dx terms: x dy/dx - (1/y) dy/dx, grouping to (x - 1/y) dy/dx. Constants: 2x + y, so dy/dx = - (2x + y) / (x - 1/y). Choice B tempts with +1/y but fails due to sign error in denominator. Recognize in logarithmic equations with polynomial terms.

Question 14

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

Explanation: This problem requires implicit differentiation for arctan(y) = x² y to find dy/dx. Differentiating gives [1/(1 + y²)] dy/dx = 2x y + x² dy/dx, chain on left, product on right. Dy/dx terms: [1/(1 + y²)] dy/dx - x² dy/dx, grouping to [1/(1 + y²) - x²] dy/dx. Equals 2x y, so dy/dx = 2x y / [1/(1 + y²) - x²]. Choice C tempts with +x² but fails due to sign error in denominator. Recognize in inverse trig functions equal to polynomials in x and y.

Question 15

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

Explanation: This problem requires implicit differentiation for x² sin y + y² cos x = 0 to find dy/dx. Differentiating gives 2x sin y + x² cos y dy/dx + 2y cos x dy/dx - y² sin x = 0, using product and chain. Dy/dx terms are x² cos y dy/dx + 2y cos x dy/dx, grouping to (x² cos y + 2y cos x) dy/dx. Constants are 2x sin y - y² sin x, so dy/dx = - (2x sin y - y² sin x) / (x² cos y + 2y cos x), but A has + y² sin x in numerator. Wait, verifying: actually, moving terms, it's - (2x sin y + y² sin x wait no: constants 2x sin y - y² sin x to other side makes -(2x sin y - y² sin x) = -2x sin y + y² sin x, but A has 2x sin y - y² sin x in numerator with negative, which is equivalent after sign. No, A is - (2x sin y - y² sin x)/denom? Wait, A has - (2x sin y - y² sin x), but let's calculate properly. Differentiating: derivative of y² cos x is 2y dy/dx cos x + y² (-sin x), so 2y cos x dy/dx - y² sin x. So total: 2x sin y + x² cos y y' + 2y cos x y' - y² sin x = 0. So y' (x² cos y + 2y cos x) = -2x sin y + y² sin x. Thus y' = (-2x sin y + y² sin x) / (x² cos y + 2y cos x) = - (2x sin y - y² sin x) / denom, which is A. Yes. Choice B tempts by flipping signs in numerator but fails to match the correct collection. Recognize in zero-equated trig products.

Question 16

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation sin(xy) + x² = y. Differentiating both sides gives cos(xy) * (y + x dy/dx) + 2x = dy/dx, incorporating the chain rule for sin(xy). The dy/dx terms appear as cos(xy) * x dy/dx on the left and -dy/dx on the right, which can be grouped as [cos(xy) x - 1] dy/dx. Rearranging yields dy/dx = (2x + y cos(xy)) / (1 - x cos(xy)), with numerator from x terms and denominator from y terms. Choice D is a tempting distractor but fails due to a sign error in the denominator, using +x cos(xy) instead of -x cos(xy) from moving terms. To recognize implicit differentiation needs, identify equations mixing trig functions of products like xy with polynomials.

Question 17

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

Explanation: This problem requires implicit differentiation for x² + sin y = y cos x to find dy/dx. Differentiating gives 2x + cos y dy/dx = dy/dx cos x - y sin x, using chain on sin y and product on y cos x. Dy/dx terms are cos y dy/dx - cos x dy/dx, grouping to (cos y - cos x) dy/dx. Constants are 2x + y sin x, so dy/dx = -(2x + y sin x)/(cos y - cos x), adjusting signs. Choice C tempts with +cos x in denominator but fails due to incorrect sign when collecting terms. Spot implicit differentiation in trig equations with x and y arguments.

Question 18

1. $\dfrac{dy}{dx}=\dfrac{6y-3x^2}{3y^2-6x}$ (correct answer)
2. $\dfrac{dy}{dx}=\dfrac{3x^2-6y}{3y^2-6x}$
3. $\dfrac{dy}{dx}=\dfrac{6y-3x^2}{3y^2}$
4. $\dfrac{dy}{dx}=\dfrac{6-3x^2}{3y^2-6x}$
5. $\dfrac{dy}{dx}=\dfrac{6y-3x^2}{3y^2+6x}$

Explanation: This problem requires implicit differentiation for the path x³ + y³ = 6xy to find dy/dx. Differentiating gives 3x² + 3y² dy/dx = 6(y + x dy/dx), applying the chain rule to y³ and product rule to xy. Dy/dx terms are 3y² dy/dx - 6x dy/dx on one side, factoring to (3y² - 6x) dy/dx, with 3x² - 6y on the other. Solving yields dy/dx = (6y - 3x²)/(3y² - 6x), simplifying by factoring out 3s where possible. Choice B tempts by swapping numerator and denominator signs but fails as it reverses the correct grouping from solving the equation. Recognize implicit differentiation when equations are symmetric in x and y, like folium of Descartes forms.

Question 19

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

Explanation: This problem uses implicit differentiation for x/y + y/x = 5 to find dy/dx. Rewrite as (x² + y²)/(xy) = 5, but directly: differentiating (x y^{-1} + y x^{-1}) = (1/y) + x (-1/y²) dy/dx + (1/x) dy/dx + y (-1/x²) = 0. Dy/dx terms: - (x/y²) dy/dx + (1/x) dy/dx, grouping to (1/x - x/y²) dy/dx. Constants: 1/y - y/x², so dy/dx = (y/x² - 1/y) / (x/y² - 1/x), but A has positive forms with signs. Wait, actually matching A after sign adjustments. Choice D tempts with + in denominator but fails by changing the sign from moving terms. Recognize in rational expressions like x/y + y/x.

Question 20

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

Explanation: This problem requires implicit differentiation for x e^y + y e^x = 1 to find dy/dx. Differentiating gives e^y + x e^y dy/dx + e^x dy/dx + y e^x = 0, product and chain rules. Dy/dx terms: x e^y dy/dx + e^x dy/dx, factoring to (x e^y + e^x) dy/dx. Constants: e^y + y e^x, so dy/dx = - (e^y + y e^x) / (x e^y + e^x). Choice B tempts with -e^x but fails by altering denominator sign. Recognize in exponential products like x e^y.