If x e^{xy} = y + sin^2 x, then the value of | vedclass

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(B) Given equation: $x e^{xy} = y + \sin^2 x$. Step $1$: Find the value of $y$ at $x = 0$. Substitute $x = 0$ into the equation: $0 \cdot e^{0} = y +

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(B) Given equation: $x e^{xy} = y + \sin^2 x$. Step $1$: Find the value of $y$ at $x = 0$. Substitute $x = 0$ into the equation: $0 \cdot e^{0} = y + \sin^2(0) \implies 0 = y + 0 \implies y = 0$. Step $2$: Differentiate both sides with respect to $x$ using the product rule and chain rule: $\frac{d}{dx}(x e^{xy}) = \frac{d}{dx}(y + \sin^2 x)$ $e^{xy} + x \cdot e^{xy} \cdot (y + x \frac{dy}{dx}) = \frac{dy}{dx} + 2 \sin x \cos x$. Step $3$: Substitute $x = 0$ and $y = 0$ into the differentiated equation: $e^{0} + 0 \cdot e^{0} \cdot (0 + 0 \cdot \frac{dy}{dx}) = \frac{dy}{dx} + 2 \sin(0) \cos(0)$ $1 + 0 = \frac{dy}{dx} + 0$ $\frac{dy}{dx} = 1$.

If $x e^{xy} = y + \sin^2 x$, then the value of $\frac{dy}{dx}$ at $x = 0$ is equal to

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