Question

If $e^y+xy=e,$ then ${\frac{d^{2}y}{d{x}^2}}_{x=0}$ is

• A. $\frac{1}{e^2}$
• B. $e^{-1}$
• C. $e$
• D. None of these

Answer 1

The correct option is A $\frac{1}{e^2}$

We have $e^y+xy=e$

Differentiating w.r.t $x$, we get

$e^y\frac{dy}{dx}+y+x\frac{dy}{dx}=0$ -----------(1)

Differentiating again w.r.t $x$, we get

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

Put $x=0$ in $e^y+xy=e$, we get $y=1$

Putting $x=0,y=1$ in (1) we get $e\frac{dy}{dx}+1=0\Rightarrow \frac{dy}{dx}=-\frac{1}{e}$

Putting $x=0,y=1,\frac{dy}{dx}=-\frac{1}{e}$ in (2) we get $\frac{d^{2}y}{d{x}^2}=\frac{1}{e^2}$