AP Calculus BC 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 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. $\dfrac{y-\frac{1}{x+y}}{\frac{1}{x+y}-x}$ (correct answer)
2. $\dfrac{\frac{1}{x+y}-y}{\frac{1}{x+y}-x}$
3. $\dfrac{y-\frac{1}{x+y}}{\frac{1}{x+y}+x}$
4. $\dfrac{y-\frac{1}{x+y}}{\frac{1}{x+y}}$
5. $\dfrac{\frac{1}{x+y}}{y}$

Explanation: This problem requires implicit differentiation to find dy/dx for the equation $\ln(x + y) = xy$. Differentiating both sides with respect to x, we treat y as a function of x, so dy/dx appears in the chain rule for $\ln(x + y)$, yielding $\frac{1}{x + y} (1 + \frac{dy}{dx})$, and in the product rule for $xy$, yielding $y + x \frac{dy}{dx}$. This gives $\frac{1}{x + y}(1 + \frac{dy}{dx}) = y + x \frac{dy}{dx}$. To isolate dy/dx, multiply both sides by $(x + y)$ or distribute: $\frac{1}{x + y} + \frac{dy}{dx} \frac{1}{x + y} = y (x + y) + x (x + y) \frac{dy}{dx}$, but better to bring all to one side and solve for $\frac{dy}{dx} = \frac{y - \frac{1}{x + y}}{\frac{1}{x + y} - x}$. A tempting distractor is choice B, $\frac{\frac{1}{x + y} - y}{\frac{1}{x + y} - x}$, which swaps signs and could come from incorrect subtraction when isolating. In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect dy/dx terms, and solve for dy/dx.

Question 2

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation x² = y cos(y). Differentiating both sides with respect to x, we treat y as a function of x, so dy/dx appears in the product rule for y cos(y), yielding cos(y) dy/dx + y (-sin(y)) dy/dx, while the left side is 2x. This gives 2x = cos(y) dy/dx - y sin(y) dy/dx. To isolate dy/dx, factor it out: 2x = dy/dx (cos(y) - y sin(y)), so dy/dx = 2x / (cos(y) - y sin(y)). A tempting distractor is choice E, 2x / (cos(y) + y sin(y)), which might arise from a sign error in the derivative of cos(y). In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect dy/dx terms, and solve for dy/dx.

Question 3

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation sin(xy) + y = x². Differentiating both sides with respect to x, we treat y as a function of x, so dy/dx appears in the chain rule for sin(xy), which is cos(xy) * (y + x dy/dx), and also in the derivative of y, which is dy/dx. This results in cos(xy)(y + x dy/dx) + dy/dx = 2x. To isolate dy/dx, expand and group terms: y cos(xy) + x cos(xy) dy/dx + dy/dx = 2x, then dy/dx (x cos(xy) + 1) = 2x - y cos(xy), so dy/dx = (2x - y cos(xy)) / (x cos(xy) + 1). A tempting distractor is choice D, (2x - y cos(xy)) / (cos(xy) + 1), which could result from forgetting the x multiplier in the denominator from the chain rule. In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect dy/dx terms, and solve for dy/dx.

Question 4

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation x² + xy + y² = 7. Differentiating both sides with respect to x, we treat y as a function of x, so dy/dx appears in the product rule for xy, yielding y + x dy/dx, and in the chain rule for y², yielding 2y dy/dx. This results in 2x + y + x dy/dx + 2y dy/dx = 0. To isolate dy/dx, group terms: x dy/dx + 2y dy/dx = -2x - y, so dy/dx (x + 2y) = - (2x + y), thus dy/dx = - (2x + y) / (x + 2y). A tempting distractor is choice D, -2x / (x + 2y), which might result from omitting the y term in the numerator during collection. In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect dy/dx terms, and solve for dy/dx.

Question 5

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation x³ + y³ = 6xy. Differentiating both sides with respect to x, we treat y as a function of x, so dy/dx appears in the chain rule for y³, yielding 3y² dy/dx, and in the product rule for 6xy, yielding 6y + 6x dy/dx. This results in 3x² + 3y² dy/dx = 6y + 6x dy/dx. To isolate dy/dx, move terms: 3y² dy/dx - 6x dy/dx = 6y - 3x², then dy/dx (3y² - 6x) = 6y - 3x², so dy/dx = (6y - 3x²) / (3y² - 6x) = (2y - x²) / (y² - 2x). A tempting distractor is choice B, (x² - 2y) / (y² - 2x), which is the negative of the correct answer and could result from a sign error in isolating terms. In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect dy/dx terms, and solve for dy/dx.

Question 6

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation e^{xy} + x = 4y, where y is defined implicitly as a function of x. Differentiating both sides with respect to x, we apply the chain and product rules to e^{xy}, yielding e^{xy}(y + x dy/dx) + 1 = 4 dy/dx. The dy/dx terms appear because y is a function of x in the exponent and the right side. To isolate dy/dx, we collect terms: e^{xy} x dy/dx - 4 dy/dx = -e^{xy} y - 1, then factor and divide to get dy/dx = (e^{xy} y + 1) / (4 - e^{xy} x). A tempting distractor like choice E changes the sign in the numerator, perhaps by mishandling the movement of terms. In general, for implicit differentiation, treat y as a function of x, apply appropriate rules to composite functions, and isolate dy/dx algebraically.

Question 7

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation ln y + x²y = 7, where y is defined implicitly as a function of x. Differentiating both sides with respect to x, we get (1/y) dy/dx + (2xy + x² dy/dx) = 0, with dy/dx appearing from the chain rule on ln y and product rule on x²y. The dy/dx terms emerge because y is treated as a function of x. To isolate dy/dx, we collect: (1/y) dy/dx + x² dy/dx = -2xy, then factor and divide to get dy/dx = -2xy / (1/y + x²). A tempting distractor like choice B omits the y in the numerator, perhaps forgetting the product rule's full application. In general, for implicit differentiation, differentiate term by term using chain and product rules, then isolate dy/dx through factoring and division.

Question 8

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation x cos y + y sin x = 0, where y is defined implicitly as a function of x. Differentiating both sides with respect to x, we get cos y - x sin y dy/dx + y cos x + sin x dy/dx = 0, with dy/dx appearing from the product and chain rules. These terms arise because y depends on x in trigonometric functions. To isolate dy/dx, we group: (-x sin y + sin x) dy/dx = -cos y - y cos x, then divide to get dy/dx = (y cos x + cos y) / (x sin y - sin x). A tempting distractor like choice D omits the cos y term, possibly by neglecting part of the differentiation. In general, for implicit differentiation involving trig functions, apply product and chain rules meticulously, then solve for dy/dx.

Question 9

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation x³ + y³ = 6xy, where y is defined implicitly as a function of x. Differentiating both sides with respect to x, we get 3x² + 3y² dy/dx = 6(y + x dy/dx), with dy/dx appearing from the chain rule on y³ and product rule on xy. These dy/dx terms arise because y depends on x. To isolate dy/dx, we rearrange: 3y² dy/dx - 6x dy/dx = 6y - 3x², then factor and divide to get dy/dx = (6y - 3x²) / (3y² - 6x). A tempting distractor like choice E reverses the signs in the numerator, possibly from incorrect subtraction. In general, for implicit differentiation, apply chain and product rules carefully to all terms, then solve the linear equation for dy/dx.

Question 10

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

Explanation: This problem requires implicit differentiation to find $\frac{dy}{dx}$ for the equation $x^2 + y^2 = 4y$, where y is defined implicitly as a function of x. Differentiating both sides with respect to x, we get $2x + 2y \frac{dy}{dx} = 4 \frac{dy}{dx}$, with $\frac{dy}{dx}$ appearing from the chain rule on $y^2$ and the right side. The $\frac{dy}{dx}$ terms appear because y is a function of x. To isolate $\frac{dy}{dx}$, we rearrange: $2y \frac{dy}{dx} - 4 \frac{dy}{dx} = -2x$, then factor and divide to get $\frac{dy}{dx} = \frac{-x}{y - 2}$. A tempting distractor like choice C doubles the denominator incorrectly, perhaps by mishandling the coefficient. In general, for implicit differentiation, differentiate each side, collect $\frac{dy}{dx}$ terms, and solve the equation for the derivative.

Question 11

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

Explanation: This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the equation $y e^x + x e^y = 10$, where $y$ is defined implicitly as a function of $x$. Differentiating both sides with respect to $x$, we get $y' e^x + y e^x + e^y + x e^y y' = 0$, with $y'$ appearing from product and chain rules. These $y'$ terms arise because $y$ depends on $x$ in the exponentials. To isolate $y'$, we group: $y' e^x + x e^y y' = -y e^x - e^y$, then factor and divide to get $y' = \dfrac{-(y e^x + e^y)}{e^x + x e^y}$. A tempting distractor like choice C omits the $y$ and $x$ factors in numerator and denominator, possibly simplifying incorrectly. In general, for implicit differentiation with exponentials, apply product and chain rules, then isolate the derivative term.

Question 12

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

Explanation: This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the equation $x^2 y + 3 y^2 = 12$, where y is defined implicitly as a function of x. Differentiating both sides with respect to x, we apply the product rule to $x^2 y$, yielding $2xy + x^2 \dfrac{dy}{dx}$, and the chain rule to $3 y^2$, yielding $6y \dfrac{dy}{dx}$. This introduces $\dfrac{dy}{dx}$ terms because y is a function of x, and the derivative of terms involving y must account for that. To isolate $\dfrac{dy}{dx}$, we collect the terms with $\dfrac{dy}{dx}$ on one side: $x^2 \dfrac{dy}{dx} + 6y \dfrac{dy}{dx} = -2xy$, then factor $\dfrac{dy}{dx}$ and divide to get $\dfrac{dy}{dx} = \frac{-2xy}{x^2 + 6y}$. A tempting distractor like choice E omits the $x^2$ term in the denominator, perhaps by forgetting the product rule on $x^2 y$. In general, for implicit differentiation, differentiate each term with respect to x, applying chain and product rules as needed, then solve algebraically for $\dfrac{dy}{dx}$.

Question 13

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation x²y + 3y² = 12. Differentiating both sides with respect to x, we treat y as a function of x, so the derivative introduces dy/dx terms via the product rule for x²y, yielding 2xy + x² dy/dx, and the chain rule for 3y², yielding 6y dy/dx. Setting the sum equal to zero gives 2xy + x² dy/dx + 6y dy/dx = 0. To isolate dy/dx, collect like terms: dy/dx (x² + 6y) = -2xy, so dy/dx = -2xy / (x² + 6y). A tempting distractor is choice C, -2xy / (2x + 6y), which might arise from incorrectly applying the product rule by using 2x instead of 2xy or mishandling factoring. In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect dy/dx terms, and solve for dy/dx.

Question 14

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

Explanation: This problem requires implicit differentiation to find dy/dx for the equation x √y + y = 10. Differentiating both sides with respect to x, we treat y as a function of x, so dy/dx appears in the product rule for x √y, yielding √y + x (1/(2√y)) dy/dx, and in the chain rule for y, yielding dy/dx. This gives √y + (x/(2√y)) dy/dx + dy/dx = 0. To isolate dy/dx, move the non-dy/dx term: (x/(2√y)) dy/dx + dy/dx = -√y, so dy/dx (x/(2√y) + 1) = -√y, thus dy/dx = -√y / (x/(2√y) + 1). A tempting distractor is choice B, -√y / (1/(2√y) + 1), which might result from forgetting the x in the derivative of x √y. In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect dy/dx terms, and solve for dy/dx.

Question 15

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

Explanation: This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the equation $e^y + xy = 5$. Differentiating both sides with respect to x, we treat y as a function of x, so $\dfrac{dy}{dx}$ appears in the chain rule for $e^y$, yielding $e^y \dfrac{dy}{dx}$, and in the product rule for $xy$, yielding $y + x \dfrac{dy}{dx}$. This gives $e^y \dfrac{dy}{dx} + y + x \dfrac{dy}{dx} = 0$. To isolate $\dfrac{dy}{dx}$, group the $\dfrac{dy}{dx}$ terms: $\dfrac{dy}{dx} (e^y + x) = -y$, so $\dfrac{dy}{dx} = \dfrac{-y}{e^y + x}$. A tempting distractor is choice B, $\dfrac{-y}{e^y}$, which might occur if one forgets to differentiate the $xy$ term properly and omits the $x \dfrac{dy}{dx}$ part. In general, when performing implicit differentiation, differentiate each term with respect to x, apply the chain rule for y terms, collect $\dfrac{dy}{dx}$ terms, and solve for $\dfrac{dy}{dx}$.