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.

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.

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.

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.

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.

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.

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.

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.

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.

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.