The curve given by x + y = e^{xy} has a tange | vedclass

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(B) Given the curve $x + y = e^{xy}$.Differentiating both sides with respect to $x$:$1 + \frac{dy}{dx} = e^{xy} \left( y + x \frac{dy}{dx} \right)$Rea

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$(1, 0)$

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(B) Given the curve $x + y = e^{xy}$.Differentiating both sides with respect to $x$:$1 + \frac{dy}{dx} = e^{xy} \left( y + x \frac{dy}{dx} \right)$Rearranging the terms to solve for $\frac{dy}{dx}$:$1 + \frac{dy}{dx} = y e^{xy} + x e^{xy} \frac{dy}{dx}$$\frac{dy}{dx} (1 - x e^{xy}) = y e^{xy} - 1$$\frac{dy}{dx} = \frac{y e^{xy} - 1}{1 - x e^{xy}}$$A$ tangent is parallel to the $y-$ axis when the slope $\frac{dy}{dx}$ is undefined,which occurs when the denominator is zero:$1 - x e^{xy} = 0$$x e^{xy} = 1$Since $e^{xy} = x + y$,we substitute this into the equation:$x(x + y) = 1$$x^2 + xy = 1$Checking the given options:For $(1, 0)$: $1^2 + (1)(0) = 1 + 0 = 1$. This satisfies the condition.Thus,the tangent is parallel to the $y-$ axis at the point $(1, 0)$.

The curve given by $x + y = e^{xy}$ has a tangent parallel to the $y-$ axis at the point

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