If {e^y} + xy = e,the ordered pair left( {fra | vedclass

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(C) Given the equation ${e^y} + xy = e$. At $x = 0$,${e^y} + 0 = e \implies {e^y} = e \implies y = 1$. Differentiating both sides with respect to $x$:

Vedclass · Accepted Answer

$\left( { - \frac{1}{e},\frac{1}{{{e^2}}}} \right)$

Vedclass · Answer

(C) Given the equation ${e^y} + xy = e$. At $x = 0$,${e^y} + 0 = e \implies {e^y} = e \implies y = 1$. Differentiating both sides with respect to $x$: ${e^y} \frac{dy}{dx} + x \frac{dy}{dx} + y = 0$. At $(0, 1)$,${e^1} \frac{dy}{dx} + 0 + 1 = 0 \implies e \frac{dy}{dx} = -1 \implies \frac{dy}{dx} = -\frac{1}{e}$. Now,differentiate ${e^y} \frac{dy}{dx} + x \frac{dy}{dx} + y = 0$ again with respect to $x$: ${e^y} \frac{d^2y}{dx^2} + {e^y} \left( \frac{dy}{dx} \right)^2 + x \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} = 0$. Substitute $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$: $e \frac{d^2y}{dx^2} + e \left( -\frac{1}{e} \right)^2 + 0 + 2 \left( -\frac{1}{e} \right) = 0$. $e \frac{d^2y}{dx^2} + \frac{1}{e} - \frac{2}{e} = 0$. $e \frac{d^2y}{dx^2} = \frac{1}{e} \implies \frac{d^2y}{dx^2} = \frac{1}{e^2}$. Thus,the ordered pair is $\left( -\frac{1}{e}, \frac{1}{e^2} \right)$.

If ${e^y} + xy = e$,the ordered pair $\left( {\frac{{dy}}{{dx}},\frac{{{d^2}y}}{{d{x^2}}}} \right)$ at $x = 0$ is equal to

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