If x{e^{xy}} = y + {sin ^2}x,then at x = 0,fr | vedclass

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(C) Given the equation $x{e^{xy}} = y + {\sin ^2}x$. When $x = 0$,substituting into the equation gives $0 \cdot {e^0} = y + {\sin ^2}(0)$,which implie

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(C) Given the equation $x{e^{xy}} = y + {\sin ^2}x$. When $x = 0$,substituting into the equation gives $0 \cdot {e^0} = y + {\sin ^2}(0)$,which implies $y = 0$. Differentiating both sides with respect to $x$ using the product rule and chain rule: $\frac{d}{dx}(x{e^{xy}}) = \frac{d}{dx}(y + {\sin ^2}x)$ ${e^{xy}} + x{e^{xy}} \cdot \frac{d}{dx}(xy) = \frac{dy}{dx} + 2\sin x \cos x$ ${e^{xy}} + x{e^{xy}} \left( y + x \frac{dy}{dx} \right) = \frac{dy}{dx} + \sin(2x)$ Substituting $x = 0$ and $y = 0$ into the differentiated equation: ${e^0} + 0 \cdot {e^0} (0 + 0 \cdot \frac{dy}{dx}) = \frac{dy}{dx} + \sin(0)$ $1 + 0 = \frac{dy}{dx} + 0$ Therefore,$\frac{dy}{dx} = 1$.

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

If $x{e^{xy}} = y + {\sin ^2}x$,then at $x = 0$,$\frac{dy}{dx} = $

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