Closed form for $n$th derivative of exponential of $f$ What is the closed form for:
$$\frac{\partial^n}{\partial x^n}\exp(f(x))=\exp(f(x))\cdot[????]$$
 A: See the Wikipedia articles titled Exponential formula and Faà di Bruno's formula.  Both involve enumerations of set partitions.
It's simpler to express with $\dfrac{\partial^n}{\partial x_1\,\partial x_2\,\cdots\,\partial x_n}$ than with $\dfrac{\partial^n}{\partial x^n}$, but after doing that, just declare $x_1,\ldots,x_n$ to be all the same variable and then drop the subscripts and collect like terms.  So for example:
\begin{align}
& \frac{\partial^3}{\partial x_1\,\partial x_2\,\partial x_3} e^{f(x)} \\[10pt]
& = e^{f(x)}\left(\frac{\partial^3 f(x)}{\partial x_1\,\partial x_2\,\partial x_3} \right. \\[10pt]
& {} + \underbrace{\frac{\partial f(x)}{\partial x_1} \cdot \frac{\partial^2 f(x)}{\partial x_2\,\partial x_3}}
+ \underbrace{\frac{\partial f(x)}{\partial x_2} \cdot \frac{\partial^2 f(x)}{\partial x_2\,\partial x_3}}
+ \underbrace{\frac{\partial f(x)}{\partial x_3} \cdot \frac{\partial^2 f(x)}{\partial x_1\,\partial x_2}} \\[10pt]
& \left.{} + \frac{\partial f(x)}{\partial x_1} \cdot \frac{\partial f(x)}{\partial x_2} \cdot \frac{\partial f(x)}{\partial x_3} \right).
\end{align}
You get one term for each partition of the set of three variables.  For the fourth derivative, there's one for each of the $15$ partitions of the set of four variables, etc.
Consequently
$$
\frac{\partial^3 f(x)}{\partial x^3} = e^{f(x)}\left( \frac{\partial^3 f(x)}{\partial x^3} + 3 \frac{\partial f(x)}{\partial x}\cdot\frac{\partial^2 f(x)}{\partial x^2} + \left( \frac{\partial f(x)}{\partial x} \right)^3  \right),
$$
and similarly with the fourth derivative the sum of the coefficients is $15$, etc.
Later note: Here's a summary of the proof: Mathematical induction on the order of the derivative.
A: So we know that if $f,g$ are infinitely differentiable, we can compute
$$
\frac{d}{dx} \left( e^{f(x)} g(x) \right) = \left( f'(x) e^{f(x)} \right) g(x) + e^{f(x)} g'(x) =  e^{f(x)} \left( f'(x) g(x) + g'(x) \right).
$$ 
So since $e^{f(x)} = e^{f(x)} \left( 1 \right)$ and that we know that all derivatives of $e^{f(x)}$ are going to be of the form $e^{f(x)}g(x)$ for some functions $g$, we can apply the above formula and assume that 
$$
\frac {d^n}{dx^n} e^{f(x)} = e^{f(x)} g_n(x), \qquad g_0(x) = 1.
$$
We can use the formula found above to find a recurrence relation for $g_n(x)$ : 
$$
g_n(x) = f'(x) g_{n-1}(x) + g_{n-1}'(x), \qquad g_0(x)
$$
The formula from Wolfram Alpha provided by Integral looks wrong because there should be no $f(x)^j$ in there ; it's impossible to make the function $f$ appear in the factor because all derivatives of $e^{f(x)}$ have the derivatives of $f$ appear in the factor (i.e. the function $g_n$), but never the function $f$ itself.
If you wanted this for yourself to compute higher derivatives of $e^{f(x)}$ I suggest you just apply the recurrence relation to get the first few. For a general formula I must admit I don't know. Would require more thinking.
A: I dont know how to get this result, but check wolfram's solution. It's exactly what you are looking for.
http://www.wolframalpha.com/input/?i=d%5En%2Fdx%5En%28e%5E%28f%28x%29%29%29
A: Related problem:(I), (II). Here is a formula for the $n$th derivative of the function $e^{\sin(x)}$

$$\left( e^{\sin(x)}\right)^{(n)}={{ i}^{{n} }}\ e^{\sin ( x )} \sum _{k=0}^{n}  
\sum _{j=0}^{n-k}\sum _{m = 0
}^{k}  \frac{\left(-1 \right)^{n-k-m} {i}^{{ j + m}} }{{2}^{(j+m)}}  {n\choose k} \left\{\matrix{n-k\\j}\right\} \left\{\matrix{k\\m}\right\}  {\rm e}^{ i x (m - j)}\\  n \in \mathbb{N},\, i = \sqrt{-1}. $$  

where $\left\{\matrix{n\\k}\right\}$ are Stirling numbers of the second kind.
Note: I'll appreciate if someone is interested in verifying this formula with Maple or Mathematica. I already tried it with Maple and it should be fine.
A: This is a simple application of Faà di Bruno's formula.
$$\partial_x^n e^{f(x)}=e^{f(x)}\sum \frac{n!}{\prod_{j=1}^n m_j!\,j!^{m_j}}\cdot \prod_{j=1}^nf^{(j)}(x)^{m_j},$$
where the sum is over all $n$-tuples $(m_1,\dots,m_n)\in{\Bbb N}^n$ such that $\sum_{j=1}^nj\,m_j=n$.
