Well, yes and no. I wanted:
\[
f'_{n+1}(x) = \frac{\exp(x)}{f'_n(f_{n+1}(x))}\\
\]
Because this is a first order ODE; and is perfectly solvable. It's simply an equation:
\[
y'_{n+1} = g_n(x,y_{n+1})\\
\]
Which is really easy to solve for locally. Once you have \(f_n\), you get \(g_n\); this completely determines \(f_{n+1}(x)\) and ensures differentiability. It produces an exotic function yes; but it's approximating a half-exponential, which is the most exotic function I can think of.
It is more so a numerical procedure than a closed form expression, I'll give you that.
\[
f'_{n+1}(x) = \frac{\exp(x)}{f'_n(f_{n+1}(x))}\\
\]
Because this is a first order ODE; and is perfectly solvable. It's simply an equation:
\[
y'_{n+1} = g_n(x,y_{n+1})\\
\]
Which is really easy to solve for locally. Once you have \(f_n\), you get \(g_n\); this completely determines \(f_{n+1}(x)\) and ensures differentiability. It produces an exotic function yes; but it's approximating a half-exponential, which is the most exotic function I can think of.
It is more so a numerical procedure than a closed form expression, I'll give you that.