A) \[\frac{1}{e}\]
B) \[\frac{1}{{{e}^{2}}}\]
C) \[\frac{1}{{{e}^{3}}}\]
D) None of these
Correct Answer: B
Solution :
We have \[{{e}^{y}}+xy=e.\]Differentiating w.r.t. x, we get \[{{e}^{y}}\frac{dy}{dx}+y+x\frac{dy}{dx}=0\] .....(i) Differentiating w.r.t. x, we get \[{{e}^{y}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{e}^{y}}{{\left( \frac{dy}{dx} \right)}^{2}}+2\frac{dy}{dx}+x\frac{{{d}^{2}}y}{d{{x}^{2}}}=0\] .....(ii) Putting \[x=0\]in \[{{e}^{y}}+xy=e,\]we get \[y=1\] Putting \[x=0,\ \ y=1\]in (i), we get \[e\frac{dy}{dx}+1=0\] Þ \[\frac{dy}{dx}=-\frac{1}{e}\] Putting \[x=0,\ y=1,\ \frac{dy}{dx}=-\frac{1}{e}\]in (ii), we get \[\therefore \].
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