If $y=e^{x+e^{x+..... to ~\infty}}$, then $\frac{d y}{d x}$ is :

$\frac{y}{y+1}$ $\frac{y}{y-1}$ $\frac{y}{1-y}$ None of these

$\frac{y}{1-y}$

$y=e^{x+y} \Rightarrow \log y=x+y$ $\Rightarrow \frac{1}{y} \frac{d y}{d x}=1+\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{y}{1-y}$ Hence (3) is correct answer.