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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the equation $e^{y} + xy = e$.
At $x = 0$,$e^{y} + 0 = e$,which implies $e^{y} = e$,so $y = 1$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(e^{y} + xy) = \frac{d}{dx}(e)$
$e^{y} \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$.
At $(x, y) = (0, 1)$:
$e^{1} \frac{dy}{dx} + 1 + 0 = 0 \implies e \frac{dy}{dx} = -1 \implies \frac{dy}{dx} = -\frac{1}{e}$.
Differentiating again with respect to $x$:
$\frac{d}{dx}(e^{y} \frac{dy}{dx} + y + x \frac{dy}{dx}) = 0$
$e^{y} (\frac{dy}{dx})^2 + e^{y} \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$
$e^{y} (\frac{dy}{dx})^2 + e^{y} \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$.
Substituting $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$:
$e(-\frac{1}{e})^2 + e \frac{d^2y}{dx^2} + 2(-\frac{1}{e}) + 0 = 0$
$\frac{1}{e} + e \frac{d^2y}{dx^2} - \frac{2}{e} = 0$
$e \frac{d^2y}{dx^2} = \frac{1}{e} \implies \frac{d^2y}{dx^2} = \frac{1}{e^2}$.
The ordered pair is $(-\frac{1}{e}, \frac{1}{e^2})$.

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

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