Question

If $e^{y}+x y=e$ the ordered pair $\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)$ 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)

$x=0 \Rightarrow y=1$
$\frac{d y}{d x}=\frac{-y}{e^{y}+x}$
$\left(\frac{d y}{d x}\right)_{(0,1)}=-\frac{1}{e}$
$\left(\frac{d^{2} y}{d x^{2}}\right)_{(0,1)}=\frac{1}{e^{2}}$