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: (A)

$e^y=xy=e$
differentiate w.r.t. x
$e^y\frac{dy}{dx}+x\frac{dy}{dx}+y=0$
$\frac{dy}{dx}\left(x+e^y\right)=-y,\frac{dy}{dx}=-\frac{1}{e}$
again differentiate w.r.t. x
$e^y.\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}.e^y.\frac{dy}{dx}+x.\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+\frac{dy}{dx}=0$
$\left(x+e^y\right)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2.e^y+2\frac{dy}{dx}=0$
$e\frac{d^{2}y}{d{x}^2}+\frac{1}{e^2}+2\left(-\frac{1}{e}\right)=0$
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{1}{e^2}$