Question

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

• A. $\left( \frac{1}{e}, \frac{1}{e^2} \right)$
• B. $\left( \frac{1}{e}, -\frac{1}{e^2} \right)$
• C. $\left( -\frac{1}{e}, -\frac{1}{e^2} \right)$
• D. $\left( -\frac{1}{e}, \frac{1}{e^2} \right)$

Answer 1

Answer: (D)

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

Answer 2

The correct answer is (D) :$\left( -\frac{1}{e}, \frac{1}{e^2} \right)$.