If $\mathrm{e}^{\mathrm{y}}+\mathrm{xy}=\mathrm{e}$, the ordered pair $\left(\frac{\mathrm{dy}}{\mathrm{dx}}, \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)$ at $x=0$ is equal to :
Correct Option: 1
$e^{y}=x y=e$
differentiate w.r.t. $\mathrm{x}$
$e^{y} \frac{d y}{d x}+x \frac{d y}{d x}+y=0$
$\frac{d y}{d x}\left(x+e^{y}\right)=-y,\left.\frac{d y}{d x}\right|_{(0,1)}=-\frac{1}{e}$
again differentiate w.r.t. $\mathrm{x}$
$e^{y} \cdot \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \cdot e^{y} \cdot \frac{d y}{d x}+x \cdot \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{d y}{d x}=0$
$\left(x+e^{y}\right) \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2} \cdot e^{y}+2 \frac{d y}{d x}=0$
$e \frac{d^{2} y}{d x^{2}}+\frac{1}{e^{2}} e+2\left(-\frac{1}{e}\right)=0$
$\therefore \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{\mathrm{e}^{2}}$
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