Iterated Derivatives of function combinations? NOTE: I will use $f^{(n)}$ to represent the $n$th derivative of the function $f$.
While working with with Taylor Series, I decided to try and find formulas to make multiple differentiation of a function easier for myself. For the sum of two functions, the iterated derivative is easy if you know the iterated derivatives of each of the two functions:
$$(f+g)^{(n)}(x)=(f^{(n)}+g^{(n)})(x)$$
The formula for the product was a little bit trickier, but I found a formula that resembles slightly the binomial theorem:
$$(f\cdot g)^{(n)}(x)=\sum_{k=0}^n \binom{n}{k}(f^{(k)}\cdot g^{(n-k)})(x)$$
However, there are three that I was not able to find a formula for:
$$(1/f)^{(n)}(x)$$
$$(f\circ g)^{(n)}(x)$$
$$(f^{-1})^{(n)}(x)$$
Does anybody know corresponding formulas for these?
 A: The Lagrange inversion formula treats $(f^{-1})^{(n)}.$$
Faà di Bruno's formula treats $(f\circ g)^{(n)}.$
In a sense, the latter is simpler if the $n$th derivative is $\dfrac{\partial^n}{\partial x_1\,\cdots\,\partial x_n}$ than if it is $\dfrac{d^n}{dx^n}.$
Start with an example:
\begin{align}
& \frac{\partial^3}{\partial x_1\,\partial x_2\,\partial x_3} f(g(x))\\[10pt]
= {} & f'(g(x_1,x_2,x_3)) \frac{\partial^3}{\partial x_1\,\partial x_2\,\partial x_3} g(x_1,x_2,x_3) & & (\text{with } f') \\[10pt]
& {} + f''(  g(x_1,x_2,x_3)) \cdot \left( \begin{array}{r}
\dfrac{\partial^2}{\partial x_1\,\partial x_2} g(x_1,x_2,x_3) \cdot \dfrac{\partial}{\partial x_3} g(x_1,x_2,x_3) \\[5pt]
{} + \dfrac{\partial^2}{\partial x_1\,\partial x_3} g(x_1,x_2,x_3) \cdot \dfrac \partial {\partial x_2} g(x_1,x_2,x_3) \\[5pt]
{} +  \dfrac{\partial^2}{\partial x_2\,\partial x_3} g(x_1,x_2,x_3) \cdot \dfrac \partial {\partial x_1} g(x_1,x_2,x_3) 
\end{array} \right) & & (\text{with } f'') \\[10pt]
& {} + f'''(g(x_1,x_2,x_3)) \frac \partial {\partial x_1} g(x_1,x_2,x_3) \cdot \frac \partial {\partial x_2} g(x_1,x_2,x_3) \cdot \frac \partial {\partial x_3} g(x_1,x_2,x_3). & & (\text{with } f''')
\end{align}
There is one term for each of the five of the partitions of the set of three variables $x_1,x_2,x_3.$ The $k$th derivative of $f$ is multiplied by an expression involving all partitions of the set of independent variables into $k$ parts.
Similarly, if we had $\dfrac{\partial^6}{\partial x_1\,\partial x_2\,\partial x_3\,\partial x_4\,\partial x_5\,\partial x_6} f(g(x_1,x_2,x_3,x_4,x_5,x_6)),$ we would have listed all $203$ partitions of the set of six independent variables.
But what if there's just one independent variable? $\dfrac{d^3}{dx^3} f(g(x)) = \text{what?}$
Then just let all three variables coalesce into one variable called $x:$
\begin{align}
& \frac{d^3}{dx^3} f(g(x)) \\[15pt]
= {} & f'(g(x)) \frac{d^3}{dx^3} g(x) \quad + \quad 3 f''(g(x)) \frac {d^2}{dx^2} g(x) \cdot \frac d {dx} g(x) \quad + \quad f'''(g(x)) \left( \frac d {dx} g(x) \right)^3
\end{align}
(where the three terms involving the second derivative of $f$ have become indistinguishable from each other and hence have become just one term with the coefficient $3$).
Faà di Bruno will also treat $(1/f)^{(n)}(x)$ by regarding it as a derivative of a composite function: $ x \mapsto f(x) \mapsto 1/f(x).$
