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

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(A) Given the equation: $xe^{xy} = y + e^{\sin 2x}$.First,find the value of $y$ when $x = 0$:Substituting $x = 0$ into the equation: $0 \cdot e^0 = y

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$-1$

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(A) Given the equation: $xe^{xy} = y + e^{\sin 2x}$.First,find the value of $y$ when $x = 0$:Substituting $x = 0$ into the equation: $0 \cdot e^0 = y + e^{\sin 0} \implies 0 = y + e^0 \implies 0 = y + 1 \implies y = -1$.Differentiating both sides with respect to $x$ using the product rule and chain rule:$\frac{d}{dx}(xe^{xy}) = \frac{d}{dx}(y + e^{\sin 2x})$$e^{xy} + x \cdot e^{xy} \cdot (y + x \frac{dy}{dx}) = \frac{dy}{dx} + e^{\sin 2x} \cdot \cos 2x \cdot 2$.Now,substitute $x = 0$ and $y = -1$ into the differentiated equation:$e^{0(-1)} + 0 \cdot e^{0(-1)} \cdot (-1 + 0 \cdot \frac{dy}{dx}) = \frac{dy}{dx} + e^{\sin 0} \cdot \cos 0 \cdot 2$$1 + 0 = \frac{dy}{dx} + 1 \cdot 1 \cdot 2$$1 = \frac{dy}{dx} + 2$$\frac{dy}{dx} = 1 - 2 = -1$.

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

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