The curve y - e^{xy} + x = 0 has a vertical t | vedclass

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(C) Given the equation of the curve: $y - e^{xy} + x = 0$. Differentiating both sides with respect to $x$: $\frac{dy}{dx} - e^{xy} \cdot (y + x \frac{

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$(1, 0)$

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(C) Given the equation of the curve: $y - e^{xy} + x = 0$. Differentiating both sides with respect to $x$: $\frac{dy}{dx} - e^{xy} \cdot (y + x \frac{dy}{dx}) + 1 = 0$. Rearranging the terms to solve for $\frac{dy}{dx}$: $\frac{dy}{dx} (1 - x e^{xy}) = y e^{xy} - 1$. $\frac{dy}{dx} = \frac{y e^{xy} - 1}{1 - x e^{xy}}$. $A$ vertical tangent occurs where the denominator is zero and the numerator is non-zero,i.e.,$\frac{dy}{dx} 	o \infty$. Setting the denominator to zero: $1 - x e^{xy} = 0$,which implies $x e^{xy} = 1$. From the original equation $y - e^{xy} + x = 0$,we have $e^{xy} = x + y$. Substituting $e^{xy} = x + y$ into $x e^{xy} = 1$,we get $x(x + y) = 1$,or $x^2 + xy = 1$. However,we must check if any point $(x, y)$ satisfies both $e^{xy} = x + y$ and $x e^{xy} = 1$. If $x = 0$,then $1 - 0 = 0$,which is $1 = 0$,a contradiction. If $x 
eq 0$,then $e^{xy} = \frac{1}{x}$. Substituting this into the original equation: $y - \frac{1}{x} + x = 0 \implies y = \frac{1}{x} - x = \frac{1 - x^2}{x}$. Now,$e^{xy} = e^{x(\frac{1-x^2}{x})} = e^{1-x^2}$. We need $e^{1-x^2} = \frac{1}{x}$. Testing the given options: For $(1, 0)$: $0 - e^0 + 1 = 0 \implies 0 - 1 + 1 = 0$ (Satisfied). At $(1, 0)$,the denominator $1 - x e^{xy} = 1 - 1 \cdot e^0 = 1 - 1 = 0$. The numerator $y e^{xy} - 1 = 0 \cdot e^0 - 1 = -1 
eq 0$. Since the denominator is $0$ and the numerator is non-zero,the slope is undefined (vertical tangent). Thus,the curve has a vertical tangent at $(1, 0)$.

The curve $y - e^{xy} + x = 0$ has a vertical tangent at

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