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.

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.

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.

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.

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

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.

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.

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.

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.

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.