Is it possible to find an expression for $\frac{d^n}{dx^n}e^{-x^2}$? I am trying to find a general form to derivatives of the function $e^{-x^2}$. I tried to do it by finding a pattern for the first derivatives but with no success. 
Any tip is welcome.

These are called Hermite polynomials, in one form or another. Let me show you show to "discover them".
Let $p_n$ be your $n$th derivative and $f=p_0$. To simplify matters I will instead consider $f= e^{x^2/2}$ (you can roll back by replacing $x$ by $\sqrt{2}xi$). Then $p_1 = x p_0$. Assume by induction that $p_n$ is of the form $q_n p_0$ for some polynomial $q_n$. Then you get that the same holds for $p_{n+1}$, since then:
$$p_{n+1} = q_n'p_0 + xq_n p_0$$
This gives you a recurrence $q_{n+1} = q_n' +x q_n$. Note that $q_0=1$ and $q_1=x$, and in fact $q_n$ is obtained by iterating the operator $\partial +x$ on $1$, that is
$$q_n = (\partial+x)^n q_0.$$
One would be tempted to use the binomial theorem here, but $\partial=d/dx$ and $x$ do not commute (!), in fact we have that for any polynomial $p$:
$\partial(xp)- x(\partial p) = p+xp'-xp' = p$
i.e. $\partial x-x \partial$ acts as the identity: we have discovered the defining relation of the Weyl algebra. This allows you to organize your computation into something that looks like a sum of terms $x^n \partial^k$ as in the binomial theorem, but now something curious happens:
applying the rule that $ \partial x=x \partial+1$ many times you get that
$$\partial^Nx = x \partial^N+N \partial^{N-1}.$$
And here comes the last (not so fun) part that you can prove by induction on $n$: it turns out that
$$(x+\partial)^n = \text{usual binomial sum} + \sum_{2a+b+c=n} \frac{n!}{a!b!c!} 2^{-a}x^b\partial ^c$$
where we only require that $c>0$. This tells us how to write down $(x+\partial)^n$ in terms of the simpler operators $x^n\partial^m$, and we know
how these act on $1$. So, for example, from
$$(x+\partial)^2  = x^2+2x\partial +\partial^2  \underline{+1}$$
you get $q_2 = x^2+1$. From
$$q_3 =  (x+\partial)^3 = x^3+3x^2\partial^2+3x\partial^2+\partial^3 \underline{+ 3x+ 3\partial}$$
you get $q_3 = x^3+3x$.
And now we can finish the computation of $q_n$: since $q_0= 1$ and since $\partial(1)=0$, we only need to consider terms in the sum above where there is no $\partial$ appearing (i.e. only take $x^n$ in the binomial sum and take only the terms with $c=0$ in the second one. This gives
$$q_n =  x^n+ \sum_{2a+b=n} \frac{(2a+b)!}{a!b!}  2^{-a}x^b.$$
You can obtain a version of the standard Hermite polynomials by changing $x$ to $2x$, so that the leading term is $2^n$ and this reads instead
$$r_n =  2^nx^n+ \sum_{2a+b=n} \frac{(2a+b)!}{a!b!}  2^{a+b}x^b.$$
The takeaway you can get here is that understanding a simple relation such as
$$\partial x-x\partial  =1$$
can give you a nice description of a concrete problem. In this case what we have done is more or less pin down how the first Weyl algebra acts on the vector space of polynomials in one variable.
A reference for the discrepancy between $(x+\partial)^n$ and the usual binomial sum in the Weyl algebra is this paper.
