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.

Set $u(x)=e^{x}$, $v(x)=-x^2$ and use the famous Faà di Bruno's Formula. 
$$
\frac{d^{n}}{d x^{n}} u(v(x))=
\sum_{m_1+m_2+\ldots+m_n=n}
\frac{n !}{m_{1} ! m_{2} ! \cdots m_{n} !} 
\cdot 
u^{\left(m_{1}+\cdots+m_{n}\right)}(v(x)) 
\cdot 
\prod_{j=1}^{n}\left(\frac{v^{(j)}(x)}{j !}\right)^{m_{j}}
$$
For $n=2$ we have
\begin{align}
\frac{d^{2}}{d x^{2}} u(v(x))
=&
\sum_{m_1+m_2=2}
\frac{2!}{m_{1} ! m_{2}!} 
\cdot 
u^{\left(m_{1}+m_{2}\right)}
\cdot 
(v(x)) 
\cdot 
\prod_{j=1}^{2}
\left(\frac{v^{(j)}(x)}{j !}\right)^{m_{j}}
\\
=&
\sum_{m_1+m_2=2}
\frac{2!}{m_{1} ! m_{2}!} 
\cdot 
u^{\left(m_{1}+m_{2}\right)}
\cdot 
(v(x)) 
\cdot 
\left(\frac{v^{(1)}(x)}{1 !}\right)^{m_{1}}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{m_{2}}
\end{align}
The partitions of $m_1+m_2=2$ are $(m_1,m_2)=(2,0),(1,1),(0,2)$. Then
\begin{align}
\frac{d^{2}}{d x^{2}} u(v(x))
=&
\frac{2!}{2 ! 0!} 
\cdot 
u^{\left(2+0\right)}
\cdot 
(v(x)) 
\cdot 
\left(\frac{v^{(1)}(x)}{1 !}\right)^{2}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{0}
\\
+&
\frac{2!}{1 ! 1!} 
\cdot 
u^{\left(1+1\right)}
\cdot 
(v(x)) 
\cdot 
\left(\frac{v^{(1)}(x)}{1 !}\right)^{1}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{1}
\\
+&
\frac{2!}{0 ! 2!} 
\cdot 
u^{\left(0+2\right)}
\cdot 
(v(x)) 
\cdot 
\left(\frac{v^{(1)}(x)}{1 !}\right)^{0}
\cdot
\left(\frac{v^{(2)}(x)}{2 !}\right)^{2}
\end{align}
A most interessant formulation is 
Faà di Bruno's determinant formula. If $u$ and $v$ are functions with a sufficient ber of derivatives, and $m \geq 1,$ then
$$
\frac{d^{m}}{d x^{m}} u(v(x))
=
\det
\left(
\begin{array}{cccccc}
\left(\begin{array}{c}
m-1 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
m-1 \\
1
\end{array}\right) u^{\prime \prime} v & \left(\begin{array}{c}
m-1 \\
2
\end{array}\right) u^{\prime \prime \prime} v & \cdots & \left(\begin{array}{c}
m-1 \\
m-2
\end{array}\right) u^{(m-1)} v & \left(\begin{array}{c}
m-1 \\
m-1
\end{array}\right) u^{(m)} v \\
-1 & \left(\begin{array}{c}
m-2 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
m-2 \\
1
\end{array}\right) u^{\prime \prime} v & \cdots & \left(\begin{array}{c}
m-2 \\
m-3
\end{array}\right) u^{(m-2)} v & \left(\begin{array}{c}
m-2 \\
m-2
\end{array}\right) u^{(m-1)} v \\
0 & -1 & \left(\begin{array}{cc}
m-3 \\
0
\end{array}\right) u^{\prime} v & \cdots & \left(\begin{array}{c}
m-3 \\
m-4
\end{array}\right) u^{(m-3)} v & \left(\begin{array}{c}
m-3 \\
m-3
\end{array}\right) u^{(m-2)} v \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & \left(\begin{array}{c}
1 \\
0
\end{array}\right) u^{\prime} v & \left(\begin{array}{c}
1 \\
1
\end{array}\right) v^{\prime \prime} v \\
0 & 0 & 0 & \cdots & -1 & \left(\begin{array}{c}
0 \\
0
\end{array}\right) u^{\prime} v
\end{array}
\right)
$$
where $u^{(i)}$ denotes $u^{(i)}(x)$ and $v^{k}$ is to be interpreted as $v^{(k)}(u(x))$
