Question

If $e^y + xy = e$, the ordered pair $\left(\frac{dy}{dx}, \frac{d^{2}y}{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: (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}{dx^{2}} + \frac{dy}{dx}. e^{y} . \frac{dy}{dx} +x. \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} + \frac{dy}{dx} = 0 $
$ \left(x+e^{y}\right) \frac{d^{2}y}{dx^{2}} + \left(\frac{dy}{dx}\right)^{2} .e^{y} + 2 \frac{dy}{dx} = 0$
$e \frac{d^2 y}{dx^2} + \frac{1}{e^2} + 2 \left( - \frac{1}{e} \right) = 0 $
$ \therefore \frac{d^{2}y}{dx^{2}} = \frac{1}{e^{2}} $