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,$\frac{dy}{dx} = 1$.

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

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