If x e^{xy} = y + sin^2 x, then frac{dy}{dx} | vedclass

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(D) Given the equation $x e^{xy} = y + \sin^2 x$. At $x = 0$, we have $0 \cdot e^{0 \cdot y} = y + \sin^2(0)$, which implies $y = 0$. Now, differentia

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(D) Given the equation $x e^{xy} = y + \sin^2 x$. At $x = 0$, we have $0 \cdot e^{0 \cdot y} = y + \sin^2(0)$, which implies $y = 0$. Now, differentiate both sides with respect to $x$: $\frac{d}{dx}(x e^{xy}) = \frac{dy}{dx} + \frac{d}{dx}(\sin^2 x)$. Using the product rule and chain rule: $x e^{xy} \left( x \frac{dy}{dx} + y \right) + e^{xy} = \frac{dy}{dx} + 2 \sin x \cos x$. Substituting $x = 0$ and $y = 0$: $0 \cdot e^0 (0 \cdot \frac{dy}{dx} + 0) + e^0 = \frac{dy}{dx} + 2 \sin(0) \cos(0)$. $0 + 1 = \frac{dy}{dx} + 0$. Therefore, $\frac{dy}{dx} = 1$ at $x = 0$.

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

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