If y=e^x + e^x + e^x+....+∞ then dy/dx=

The correct answer is Option (3) → $\frac{y}{1-y}$

$y=e^{{x + e}^{{x + e}^{{x+e}^{...+∞}}}}=y=e^{x+y}$

$\log y = x+y$

differentiating wrt x

$\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}$

so $\frac{dy}{dx}=\frac{y}{1-y}$