[b]6.[/b] Let $f(x)$ be an arbitrary function, differentiable infinitely many times. Then the $n$th derivative of $f(e^{x})$ has the form

$\frac{d^{n}}{dx^{n}}f(e^{x})= \sum_{k=0}^{n} a_{kn}e^{kx}f^{(k)}(e^{x})$ ($n=0,1,2,\dots$).

From the coefficients $a_{kn}$ compose the sequence of polynomials

$P_{n}(x)= \sum_{k=0}^{n} a_{kn}x^{k}$ ($n=0,1,2,\dots$)

and find a closed form for the function

$F(t,x) = \sum_{n=0}^{\infty} \frac{P_{n}(x)}{n!}t^{n}.$

[b](S. 22)[/b]
1. We start by considering the function \( f(e^x) \) and its \( n \)-th derivative. According to the problem, the \( n \)-th derivative of \( f(e^x) \) can be expressed as:
   \[
   \frac{d^n}{dx^n} f(e^x) = \sum_{k=0}^{n} a_{kn} e^{kx} f^{(k)}(e^x)
   \]
   where \( a_{kn} \) are coefficients that we need to determine.

2. To find these coefficients, we use the function \( f(x) = e^x \). Plugging this into the given formula, we get:
   \[
   \frac{d^n}{dx^n} e^{e^x} = \sum_{k=0}^{n} a_{kn} e^{kx} e^{e^x}
   \]
   Simplifying, we have:
   \[
   \frac{d^n}{dx^n} e^{e^x} = e^{e^x} \sum_{k=0}^{n} a_{kn} e^{kx}
   \]

3. Next, we use the Taylor expansion of \( e^{e^{x+t}} \):
   \[
   e^{e^{x+t}} = \sum_{n \geq 0} \frac{1}{n!} \left( \frac{d^n}{dx^n} e^{e^x} \right) t^n
   \]
   Substituting the expression for \( \frac{d^n}{dx^n} e^{e^x} \) from step 2, we get:
   \[
   e^{e^{x+t}} = \sum_{n \geq 0} \frac{1}{n!} \left( e^{e^x} \sum_{k=0}^{n} a_{kn} e^{kx} \right) t^n
   \]
   Simplifying further, we have:
   \[
   e^{e^{x+t}} = e^{e^x} \sum_{n \geq 0} \frac{1}{n!} \left( \sum_{k=0}^{n} a_{kn} e^{kx} \right) t^n
   \]

4. We recognize that the expression inside the sum can be written in terms of the polynomials \( P_n(x) \):
   \[
   P_n(x) = \sum_{k=0}^{n} a_{kn} x^k
   \]
   Therefore, we can rewrite the series as:
   \[
   e^{e^{x+t}} = e^{e^x} \sum_{n \geq 0} \frac{P_n(e^x)}{n!} t^n
   \]

5. Comparing both sides of the equation, we see that:
   \[
   e^{e^{x+t}} = e^{e^x} F(t, e^x)
   \]
   where:
   \[
   F(t, x) = \sum_{n=0}^{\infty} \frac{P_n(x)}{n!} t^n
   \]

6. To find the closed form of \( F(t, x) \), we note that:
   \[
   e^{e^{x+t}} = e^{e^x} e^{e^x (e^t - 1)}
   \]
   Therefore:
   \[
   F(t, x) = e^{x(e^t - 1)}
   \]

The final answer is \( \boxed{ F(t, x) = e^{x(e^t - 1)} } \).