Question

lf $xy=e-e^y$ then $\frac{d^{2}y}{d{x}^2}$, when $x=0$, is equal to

• A. $\frac{1}{e}$
• B. $\frac{1}{e^3}$
• C. $\frac{1}{e^2}$
• D. $0$

Answer 1

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

To find ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=0}=?$

$xy=e-e^y$

$\Rightarrow x\cdot \frac{dy}{dx}+y=-e^y\left(\frac{dy}{dx}\right)$ .. Taking derivative

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

$\Rightarrow \frac{dy}{dx}=\frac{-y}{(x+e^y)}$ ....... $\left(i\right)$

Differentiating both side w.r.t $x$

$\Rightarrow (1+e^y\cdot \frac{dy}{dx})\frac{dy}{dx}+(x+e^y)\frac{d^{2}y}{d{x}^2}=\frac{-dy}{dx}$ ...... $\left(ii\right)$

When $x=0$

$e-e^y=0$

$\Rightarrow e^y=e$

$\Rightarrow y=1$

When $x=0,\frac{dy}{dx}=\frac{-1}{0+e}$

$\frac{dy}{dx}=-\frac{1}{e}$

So from eqn. $\left(ii\right)$

$[1+e\cdot (-1/e)]\times (-\frac{1}{e})+(0+e^1)\frac{d^{2}y}{d{x}^2}=\frac{1}{e}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{e^2}$