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What's the general expression of $\dfrac{d ^n e^{f(x)}}{d x^n}$ in term of derivative $\dfrac{d f(x)}{d x}$, $\dfrac{d ^2e^{f(x)}}{d x^2}$,$\dfrac{d ^3e^{f(x)}}{d x^3},\dots$?

Actually I wonder whether there is a special expression for the series of coefficients for different integers $n$? Some of the series: $(1),(1,1),(1,3,1),(1,6,4,3,1),(1,10,10,5,15,10,1)\dots$

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This can help you. – Davide Giraudo Jul 18 '12 at 20:21
Why are you using partial derivative notation? Is your function a function of several variables, or just one? – Arturo Magidin Jul 18 '12 at 20:23
@Arturo Magidin sorry for my misleading expression;i've already edited my question. Here i only consider the one variable case.( the original expression is copied from Mathematica, so it is in term of partial derivative) – Mathieu Jul 18 '12 at 20:35

This is stated in the language of power series. The coefficients in the power series are the derivatives evalutated at $0$. You can do this at points other than $0$ by looking at powers of $x-x_0$ rather than powers of $x$.

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Using Faà di Bruno's formula, we can write $$\frac{d^ne^{f(x)}}{dx^n}=e^{f(x)}\sum_{k=0}^nB_{n,k}(f'(x),\dots,f^{(n-k+1)}(x)),$$ where $B_{n,k}$ is Bell's polynomial.

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From here (under the section "Exp from function"), we have the formula

$$\frac{\mathrm d^k}{\mathrm dz^k}\exp(g(z))=\exp(g(z))\sum_{m=0}^k\frac1{m!}\sum_{j=0}^m (-1)^j \binom{m}{j} g(z)^j\frac{\mathrm d^k}{\mathrm dz^k}g(z)^{m-j}$$

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