Given equation

The relation is $e^y + xy = e.$

We need to find $\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)$ at $x=0$.

Find $y$ at $x=0$

Put $x=0$ in the equation: $e^y + 0\cdot y = e$ $e^y = e$ $y=1.$

Differentiate once implicitly

Starting from $e^y + xy = e,$ differentiate with respect to $x$: $e^y\frac{dy}{dx} + x\frac{dy}{dx} + y = 0.$

So, $\left(e^y + x\right)\frac{dy}{dx} + y = 0$ $\frac{dy}{dx} = -\frac{y}{e^y + x}.$

At $x=0$, $y=1$: $\frac{dy}{dx} = -\frac{1}{e^1+0} = -\frac{1}{e}.$

Differentiate again

From $e^y\frac{dy}{dx} + x\frac{dy}{dx} + y = 0,$ differentiate again:

• Derivative of $e^y\dfrac{dy}{dx}$ is $e^y\left(\frac{dy}{dx}\right)^2 + e^y\frac{d^2y}{dx^2}$
• Derivative of $x\dfrac{dy}{dx}$ is $x\frac{d^2y}{dx^2} + \frac{dy}{dx}$
• Derivative of $y$ is $\frac{dy}{dx}$

Therefore, $e^y\left(\frac{dy}{dx}\right)^2 + e^y\frac{d^2y}{dx^2} + x\frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} = 0.$

Rearranging, $\left(e^y + x\right)\frac{d^2y}{dx^2} + e^y\left(\frac{dy}{dx}\right)^2 + 2\frac{dy}{dx} = 0.$

Hence, $\frac{d^2y}{dx^2} = -\frac{e^y\left(\dfrac{dy}{dx}\right)^2 + 2\dfrac{dy}{dx}}{e^y+x}.$

Substitute $x=0$, $y=1$, $\dfrac{dy}{dx}=-\dfrac{1}{e}$

$\frac{d^2y}{dx^2} = -\frac{e\left(\dfrac{1}{e^2}\right) + 2\left(-\dfrac{1}{e}\right)}{e}$ $= -\frac{\dfrac{1}{e} - \dfrac{2}{e}}{e}$ $= -\frac{-\dfrac{1}{e}}{e}$ $= \frac{1}{e^2}.$

Ordered pair

Therefore, $\left(\frac{dy}{dx},\frac{d^2y}{dx^2}\right)=\left(-\frac{1}{e},\frac{1}{e^2}\right).$

Compare with options

This matches Option B.

Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.