Question

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

• 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)

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