Question

If $e^y+xy=e,$ then the value of $\frac{d^{2}y}{d{x}^2}forx=0$is

• A. $\frac{1}{e}$
• B. $\frac{1}{e^2}$
• C. $\frac{1}{e^3}$
• D. none of these

Answer 1

The correct option is B

$\frac{1}{e^2}$

Explanation for the correct option:

Step 1: Differentiate with respect to $x$

Given $e^y+xy=e,$

When $x=0,y=1$

$e^y\frac{dy}{dx}+x\frac{dy}{dx}+y=0\dots \left(i\right)$

Put $x=0$ and $y=1$, we get

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{-y}{x+e^y}\\ & =& \frac{-1}{e}\\ & & \end{array}$

Step 2: Differentiate $\left(i\right)$ again with respect to $x$

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

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}[e^y+x]=-2\frac{dy}{dx}–e^y{\left(\frac{dy}{dx}\right)}^2\dots \left(ii\right)$

Step 3: Put $x=0,y=1$ and $\frac{dy}{dx}=-\frac{1}{e}$in $\left(ii\right)$

$$\begin{gathered} \Rightarrow \frac{d^{2}y}{d{x}^2}[e+0]=\frac{2}{e}–\frac{e}{e^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}e=\frac{2}{e}–\frac{1}{e} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}e=\frac{1}{e} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\mathbf{}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{2}}} \end{gathered}$$

Hence, Option ‘B’ is Correct.