This problem requires implicit differentiation to find dy/dx for the relation x cos y + y sin x = 3. Chain and product rules introduce dy/dx for trigonometric terms with y. Differentiation yields cos y - x sin y dy/dx + sin x dy/dx + y cos x = 0. Terms are grouped by isolating dy/dx factors, giving (-cos y - y cos x)/(-x sin y + sin x). Choice B fails as a distractor with an incorrect sign in the denominator, altering the expression. Recognize this technique for implicit trig relations where x and y are arguments or coefficients in sine and cosine.

This problem requires implicit differentiation to find dy/dx for the implicitly defined relation x² + x y + sin y = 0. When differentiating, terms like sin y produce cos y dy/dx, and x y yield x dy/dx + y. These dy/dx terms appear because y is a function of x, applying chain and product rules. To solve, group dy/dx terms (x dy/dx + cos y dy/dx) and constants (-2x - y), then isolate dy/dx. A tempting distractor like choice E fails because it leaves dy/dx unsolved in the numerator. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.

This problem requires implicit differentiation to find dy/dx for the relation x² y = sin y. Differentiating: 2 x y + x² dy/dx = cos y dy/dx. Dy/dx from product and trig chain. Grouping: 2 x y = dy/dx (cos y - x²), dy/dx = 2 x y / (cos y - x²) = - 2 x y / (x² - cos y), choice A. Choice B has + cos y, denominator error. Spot in polynomial-trig relations.

This problem requires implicit differentiation to find dy/dx for the relation x² $e^y$ + y = 4. Differentiating both sides with respect to x introduces dy/dx via the chain rule for terms involving y. The term x² $e^y$ requires the product rule, yielding 2x $e^y$ + x² $e^y$ dy/dx, while the standalone y differentiates to dy/dx. Collecting like terms groups the dy/dx factors together as (x² $e^y$ + 1) dy/dx = -2x $e^y$, allowing us to solve for dy/dx = -2x $e^y$ / (x² $e^y$ + 1). A tempting distractor like choice D forgets the dy/dx from the y term, incorrectly omitting the +1 in the denominator. To recognize implicit differentiation opportunities, look for equations defining y implicitly without solving for y explicitly.

This problem requires implicit differentiation to find dy/dx for the relation x² y + tan(x y) = 0. Differentiating: 2 x y + x² dy/dx + sec²(x y) (y + x dy/dx) = 0. Dy/dx from product and chain rules. Grouping: 2 x y + y sec²(x y) + dy/dx (x² + x sec²(x y)) = 0, dy/dx = - (2 x y + y sec²) / (x² + x sec²) = - y (2 x + sec²) / [x (x + sec²)], but simplified to choice A. Choice B has positive numerator, from sign error. Spot in trig functions of products.

This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the relation $x^2 + y^2 = \dfrac{1}{xy}$. Differentiating: $2x + 2y \dfrac{dy}{dx} = -\dfrac{1}{(xy)^2} (y + x \dfrac{dy}{dx})$. Expand: right = $- \dfrac{y + x \dfrac{dy}{dx}}{x^2 y^2}$. So $2x + 2y \dfrac{dy}{dx} + \dfrac{y + x \dfrac{dy}{dx}}{x^2 y^2} = 0$. To combine, multiply through by $x^2 y^2$: $2x \cdot x^2 y^2 + 2y \dfrac{dy}{dx} x^2 y^2 + y + x \dfrac{dy}{dx} = 0$? Wait, better: the equation is $2x + 2y y' = - \dfrac{1}{x^2 y^2} (y + x y')$. Let's group: $2x + 2y y' + \dfrac{y}{x^2 y^2} + \dfrac{x y'}{x^2 y^2} = 0$. Simplify: $2x + \dfrac{1}{x^2 y} + y' (2y + \dfrac{1}{x y^2}) = 0$. Wait, $\dfrac{y}{x^2 y^2} = \dfrac{1}{x^2 y}$, $\dfrac{x}{x^2 y^2} = \dfrac{1}{x y^2}$. Yes, so $\dfrac{dy}{dx} = - \dfrac{2x + \dfrac{1}{x^2 y}}{2y + \dfrac{1}{x y^2}}$, which is choice C. Choice A has wrong signs. Recognize in reciprocal relations.

This problem requires implicit differentiation to find dy/dx for the implicitly defined relation x² + tan y = 3y. When differentiating, terms like tan y produce sec² y dy/dx via the chain rule, and 3y yields 3 dy/dx. These dy/dx terms appear because y is a function of x, necessitating the chain rule. To solve, group dy/dx terms (sec² y dy/dx - 3 dy/dx) and constants (-2x), then isolate dy/dx. A tempting distractor like choice D fails because it leaves dy/dx unsolved in the numerator. To recognize when to use implicit differentiation, look for equations where y is not explicitly solved for in terms of x, particularly those with mixed x and y terms.

This problem requires implicit differentiation of √x + √y = 5 to find dy/dx. Rewriting as x^(1/2) + y^(1/2) = 5 and differentiating gives (1/2)x^(-1/2) + (1/2)y^(-1/2)(dy/dx) = 0. Multiplying through by 2 to clear fractions: x^(-1/2) + y^(-1/2)(dy/dx) = 0, so y^(-1/2)(dy/dx) = -x^(-1/2). Therefore, dy/dx = -x^(-1/2)/y^(-1/2) = -y^(1/2)/x^(1/2) = -√y/√x. A common error is to invert the fraction incorrectly, getting -√x/√y instead. The key insight is that when dividing by y^(-1/2), you multiply by y^(1/2), and the pattern x^(-1/2)/y^(-1/2) = y^(1/2)/x^(1/2) follows from the rule for dividing powers.

This problem uses implicit differentiation on x·cos(y) + y·sin(x) = 1 to find dy/dx. Differentiating requires the product rule on both terms: d/dx[x·cos(y)] = cos(y) + x·(-sin(y))·(dy/dx) and d/dx[y·sin(x)] = (dy/dx)·sin(x) + y·cos(x). Setting the sum equal to 0: cos(y) - x·sin(y)·(dy/dx) + (dy/dx)·sin(x) + y·cos(x) = 0. Collecting dy/dx terms: (sin(x) - x·sin(y))·(dy/dx) = -(cos(y) + y·cos(x)), so dy/dx = -(cos(y) + y·cos(x))/(sin(x) - x·sin(y)). A common error is forgetting the negative sign when cos(y) appears in the derivative of cos(y), leading to sign errors. The recognition strategy is to carefully track signs through trigonometric derivatives and note that the denominator can be written as -(x·sin(y) - sin(x)) = -x·sin(y) + sin(x).

This problem involves implicit differentiation of x² + tan y = xy. Differentiating both sides: 2x + sec²y·dy/dx = x·dy/dx + y (using chain rule for tan y and product rule for xy). Rearranging to collect dy/dx terms: sec²y·dy/dx - x·dy/dx = y - 2x, which factors as dy/dx(sec²y - x) = y - 2x. Therefore, dy/dx = (y - 2x)/(sec²y - x), but multiplying numerator and denominator by -1 gives (2x - y)/(x - sec²y), which is choice B. Choice A has the signs reversed incorrectly. The recognition strategy is to remember that d/dx[tan y] = sec²y·dy/dx and to carefully track signs when rearranging terms.