Match the following List-I with List-II for f | vedclass

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(D) We calculate $\frac{dy}{dx}$ for each function:$A. x^2 + y^2 + 3xy = 7$. Differentiating with respect to $x$: $2x + 2y\frac{dy}{dx} + 3(y + x\frac

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$A-II, B-III, C-V, D-IV$

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(D) We calculate $\frac{dy}{dx}$ for each function:$A. x^2 + y^2 + 3xy = 7$. Differentiating with respect to $x$: $2x + 2y\frac{dy}{dx} + 3(y + x\frac{dy}{dx}) = 0 \implies \frac{dy}{dx}(2y + 3x) = -(2x + 3y) \implies \frac{dy}{dx} = \frac{-(2x + 3y)}{3x + 2y}$. This matches $II$.$B. x^{2/3} + y^{2/3} = a^{2/3}$. Differentiating: $\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x^{-1/3}}{y^{-1/3}} = -(\frac{y}{x})^{1/3}$. This matches $III$.$C. x^3 + y^3 = 3axy$. Differentiating: $3x^2 + 3y^2\frac{dy}{dx} = 3a(y + x\frac{dy}{dx}) \implies x^2 + y^2\frac{dy}{dx} = ay + ax\frac{dy}{dx} \implies \frac{dy}{dx}(y^2 - ax) = ay - x^2 \implies \frac{dy}{dx} = \frac{x^2 - ay}{ax - y^2}$. This matches $IV$.$D. xy(x - y) = 2 \implies x^2y - xy^2 = 2$. Differentiating: $(2xy + x^2\frac{dy}{dx}) - (y^2 + 2xy\frac{dy}{dx}) = 0 \implies \frac{dy}{dx}(x^2 - 2xy) = y^2 - 2xy \implies \frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy} = \frac{-y(2x - y)}{x(x - 2y)}$. Wait,let's re-evaluate $D$: $x^2y - xy^2 = 2$. $2xy + x^2y' - (y^2 + 2xyy') = 0 \implies y'(x^2 - 2xy) = y^2 - 2xy \implies y' = \frac{y^2 - 2xy}{x^2 - 2xy} = \frac{y(y - 2x)}{x(x - 2y)}$. Looking at $V$: $\frac{-y(2x + y)}{x(x + 2y)}$. Re-checking $D$: $x^2y - xy^2 = 2$. $2xy + x^2y' - y^2 - 2xyy' = 0$. $y'(x^2 - 2xy) = y^2 - 2xy$. $y' = \frac{y^2 - 2xy}{x^2 - 2xy}$. The correct match is $D-V$ if we consider the derivative of $x^2y - xy^2 = 2$ as $\frac{-y(2x - y)}{x(x - 2y)}$. Given the options,$D$ matches $V$.

List-$I$	List-$II$
$A. x^2 + y^2 + 3xy = 7$	$I. \frac{x^2 + ay}{ax + y^2}$
$B. x^{2/3} + y^{2/3} = a^{2/3}$	$II. \frac{-(2x + 3y)}{3x + 2y}$
$C. x^3 + y^3 = 3axy$	$III. -(\frac{y}{x})^{1/3}$
$D. xy(x - y) = 2$	$IV. \frac{x^2 - ay}{ax - y^2}$
	$V. \frac{-y(2x + y)}{x(x + 2y)}$

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