TITLE: Computing $\frac{d^k}{dx^k}\left(f(x)^k\right)$ where $k$ is a positive integer 
QUESTION [10 upvotes]: Does anyone know a formula for the derivative $$\frac{d^k}{dx^k}\left(f(x)^k\right)$$ where $k$ is some positive integer?
I started trying to work it out but it got messy.

REPLY [2 votes]: After Raymond's explication of Davide's Faa di Bruno-formula I decomposed it in an even more memorizable form. Example  k=4
$$ \begin{eqnarray}  {(f^4)^{(4)}\over 4! \cdot 4! } &=&
  {{ f^{(0)}\over 0!} { f^{(0)}\over 0!}{ f^{(0)}\over 0!} \over 3! }{{ f^{(4)}\over 4!}  \over 1! }  \\ \\
 &+ & {{ f^{(0)}\over 0!} { f^{(0)}\over 0!} \over 2! }{{ f^{(1)}\over 1!}  \over 1! } {{ f^{(3)}\over 3!}  \over 1! } \\ \\
 &+ & {{ f^{(0)}\over 0!} { f^{(0)}\over 0!} \over 2! }{{ f^{(2)}\over 2!}{ f^{(2)}\over 2!}  \over 2! } 
 &+ & {{ f^{(0)}\over 0!}  \over 1! } {{ f^{(1)}\over 1!} { f^{(1)}\over 1!} \over 2! }{{ f^{(2)}\over 2!}  \over 1! } \\ \\
 &+ & {{ f^{(1)}\over 1!} { f^{(1)}\over 1!} { f^{(1)}\over 1!}  { f^{(1)}\over 1!}\over 4! } \\ \\
 \end{eqnarray} $$
The key for that decomposition is, that for some k we need k factors in each summand. Then the sum of all derivation-indices (with their multiplicity!) must also equal k. A j-fold multiplicity of a factor must be compensated by a j! in the denominator.
If we write 
$\displaystyle  [k]= { f^{(k)}\over k! } $ and $ \displaystyle [k \times j] = { [k]^j \over j!} $  we can denote it even more compact:
$$ \begin{eqnarray}{(f^4)^{(4)} \over 4!4! }&=& [0 \times 3][4 \times 1] \\ &+& [0 \times 2][1 \times 1][3 \times 1] \\ &+& [0 \times 2][2 \times 2] + [0 \times 1][1 \times 2][2 \times 1] \\
&+&[1 \times 4] \end{eqnarray}
$$ where, if we take the brackets as products and the terms as sums then each term has the sum k. It is a simple mechanic to write down the terms by hand for some small k, even for $k=10$ or so this should be doable with paper and pen.