This problem requires implicit differentiation to find dy/dx for the relation $e^{x+y}$ + x²y = 5. Dy/dx emerges from the chain rule applied to exponential and product terms involving y. The derivative of $e^{x+y}$ is $e^{x+y}$(1 + dy/dx), and x²y gives 2xy + x² dy/dx, summing to zero. We conceptually group dy/dx coefficients together, solving as $(-e^{x+y}$ - $2xy)/(e^{x+y}$ + x²). Choice C is a tempting distractor but omits the x² in the denominator, likely from forgetting the product rule's second part. Recognize implicit differentiation when exponential or polynomial mixtures of x and y prevent easy isolation of y.

This problem requires implicit differentiation of √(x + y) + y² = 5. The square root term uses the chain rule: d/dx[√(x + y)] = 1/(2√(x + y))·(1 + dy/dx). The complete differentiation yields: (1/(2√(x + y)))(1 + dy/dx) + 2y(dy/dx) = 0. Multiplying through by 2√(x + y) to clear the fraction: 1 + dy/dx + 4y√(x + y)(dy/dx) = 0. Factoring out dy/dx: 1 + dy/dx[1 + 4y√(x + y)] = 0, which gives dy/dx = -1/(1 + 4y√(x + y)). Choice C incorrectly places dy/dx in the denominator alongside the other terms. The key insight is to clear fractions before collecting dy/dx terms to avoid algebraic errors.

This problem requires implicit differentiation of x² + xy + y² = 7. Differentiating term by term: 2x + (y + x(dy/dx)) + 2y(dy/dx) = 0, where the middle term uses the product rule. Collecting dy/dx terms gives x(dy/dx) + 2y(dy/dx) = -2x - y, which factors as (x + 2y)(dy/dx) = -2x - y. Therefore, dy/dx = (-2x - y)/(x + 2y). Choice C shows a common algebraic error of leaving dy/dx in the denominator instead of factoring it out. The key recognition pattern is that every term containing y will contribute a dy/dx term when differentiated.

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 of sin(xy) + y = x. When differentiating sin(xy), we must use the chain rule combined with the product rule, giving cos(xy)·(y + x(dy/dx)). The full differentiation yields cos(xy)·y + cos(xy)·x(dy/dx) + dy/dx = 1. Collecting dy/dx terms gives cos(xy)·x + 1 = 1 - y·cos(xy). Therefore, dy/dx = (1 - y·cos(xy))/(x·cos(xy) + 1). Choice D incorrectly places dy/dx in the denominator rather than factoring it out. The key insight is recognizing that differentiating composite functions like sin(xy) requires both chain and product rules.

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.