Question

Let $y=e^{x+e^{x+.....\infty }}$, then find $\frac{dy}{dx}$.

Answer 1

Given

$y=e^{x+e^x+...\infty }$

Or $y=e^{x+y}$

$\log y=(x+y)$

Now differentiating both sides w.r.to $x$ we get

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

$(\frac{1}{y}-1)\frac{dy}{dx}=1$

$\frac{dy}{dx}=\left(\frac{y}{1-y}\right)$