This problem requires implicit differentiation of x²y² + x = 1. For x²y², we apply the product rule treating it as (x²)(y²): 2x·y² + x²·2y(dy/dx), and differentiating x gives 1. The equation becomes 2xy² + 2x²y(dy/dx) + 1 = 0. Solving for dy/dx: 2x²y(dy/dx) = -2xy² - 1, which gives dy/dx = -(2xy² + 1)/(2x²y). Choice B incorrectly omits the 1 in the numerator that comes from differentiating the x term on the left side. When performing implicit differentiation, remember to differentiate every term in the equation, including standalone x terms that don't contain y.

This problem asks for dy/dx from the implicit equation x² + xy + y² = 9, a conic section equation. Differentiating term by term: 2x + y + x(dy/dx) + 2y(dy/dx) = 0, where the xy term requires the product rule. Collecting dy/dx terms: x(dy/dx) + 2y(dy/dx) = -2x - y, which factors as dy/dx(x + 2y) = -(2x + y). Choice C incorrectly omits the y term in the numerator, likely from forgetting that d/dx(xy) = y + x(dy/dx), not just x(dy/dx). The recognition strategy is to always apply the product rule completely to mixed terms like xy, remembering that both factors contribute to the derivative.

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 asks for dy/dx from x + y ln x = 6, where y is multiplied by ln x. Differentiating: 1 + (ln x)(dy/dx) + y·(1/x) = 0, using the product rule on y ln x and remembering that d/dx(ln x) = 1/x. Solving for dy/dx: (ln x)(dy/dx) = -1 - y/x, so dy/dx = -(1 + y/x)/(ln x). Choice D incorrectly writes the numerator as (1 + y) instead of (1 + y/x), missing that the derivative of ln x contributes a factor of 1/x to the y term. The strategy for logarithmic products is to carefully apply the product rule, remembering that d/dx(ln x) = 1/x, not just 1.

This problem requires implicit differentiation of x sin y + y cos x = 2, involving products of x and y with trigonometric functions. Differentiating x sin y requires the product rule: sin y + x cos y(dy/dx), while differentiating y cos x gives cos x(dy/dx) - y sin x. Setting the total derivative to 0: sin y + x cos y(dy/dx) + cos x(dy/dx) - y sin x = 0, which rearranges to (x cos y + cos x)(dy/dx) = -sin y + y sin x. The common mistake in choice D is getting the wrong sign on y sin x, treating it as if it adds to sin y rather than subtracts. The key insight is that when differentiating y cos x, the derivative of cos x is -sin x, introducing a negative sign that makes the final term +y sin x in our rearrangement.

This problem requires implicit differentiation to find dy/dx from the equation x²y + y³ = 7. When we differentiate both sides with respect to x, the left side requires the product rule on x²y (giving 2xy + x²(dy/dx)) and the chain rule on y³ (giving 3y²(dy/dx)), while the right side gives 0. Collecting all terms with dy/dx on one side gives us x²(dy/dx) + 3y²(dy/dx) = -2xy, which factors as (x² + 3y²)(dy/dx) = -2xy. A common error is forgetting the product rule on x²y and only getting x²(dy/dx), which would lead to the incorrect answer B. The key strategy is to remember that every y term generates a dy/dx factor when differentiated, and to carefully apply the product rule to mixed terms like x²y.

This problem asks us to find dy/dx for the implicit relation x³ + y³ = 3xy, known as the folium of Descartes. Differentiating both sides: 3x² + 3y²(dy/dx) = 3y + 3x(dy/dx), where the right side uses the product rule on 3xy. Rearranging to collect dy/dx terms: 3y²(dy/dx) - 3x(dy/dx) = 3y - 3x², which factors as dy/dx(3y² - 3x) = 3y - 3x². The incorrect answer B reverses the signs in the denominator, getting 3y² - 3x instead of 3x - 3y² after moving terms. Remember that when moving a dy/dx term from right to left, its coefficient changes sign, so 3x(dy/dx) on the right becomes -3x(dy/dx) on the left, leading to the correct denominator 3x - 3y².

This problem requires implicit differentiation to find dy/dx for the relation where y is not explicitly a function of x. Differentiating $x^2$ tan y + y = 3, $x^2$ tan y gives $x^2$ $sec^2$ y dy/dx + 2 x tan y by product and chain, + dy/dx. Dy/dx terms from chain on tan y and standalone y. Group: 2 x tan y + $x^2$ $sec^2$ y dy/dx + dy/dx = 0, $(x^2$ $sec^2$ y + 1) dy/dx = - 2 x tan y, dy/dx = - 2 x tan y / $(x^2$ $sec^2$ y + 1). A tempting distractor like choice C fails by omitting the +1 in the denominator, forgetting the dy/dx from y. Identify implicit differentiation when trig functions of y are multiplied by x powers.

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. Note the equation is $x^2$ + $y^2$ + 2 x $y^2$ = 9, diff: 2x + 2y y' + 2 (x * 2 y y' + $y^2$) = 0, since 2 x $y^2$ by product: 2 [ $y^2$ + x * 2 y y' ]. Dy/dx appears in $y^2$ term by chain and in 2 x $y^2$ by product and chain. Group: 2x + 2y y' + 2 $y^2$ + 4 x y y' = 0, so (2y + 4 x y) y' = - (2x + 2 $y^2$), y' = - (2x + 2 $y^2$) / (2y + 4 x y) = - (x + $y^2$) / (y + 2 x y), but can factor 2: yes, matches B after simplifying by 2. A tempting distractor like choice A fails by not accounting for the extra $y^2$ terms from differentiating 2 x $y^2$ properly. Recognize implicit differentiation when polynomials involve mixed x y terms requiring product rule.