If ey+xy=e, then the value of d2ydx2 for x=0, is

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

1. A

$\frac{1}{e}$

2. B

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

3. C

$\frac{1}{{e}^{3}}$

4. D

0

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### Solution:

We have ${e}^{y}+xy=e$. Differentiating w.r.t.x, we get ${e}^{y}\frac{dy}{dx}+y+x\frac{dy}{dx}=0.....\left(i\right)$

Differentiating w.r.t.x, we get ${e}^{y}\frac{{d}^{2}y}{d{x}^{2}}+{e}^{y}{\left(\frac{dy}{dx}\right)}^{2}+2\frac{dy}{dx}+x\frac{{d}^{2}y}{d{x}^{2}}=0.....\left(ii\right)$

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

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