Consider roots $f = 0$ of a nicely-behaved real function $f(x, t)$ of two (real) variables.
Namely, points $(x, t)$ on which $f$ vanishes, $f(x, t) = 0$.
Suppose that $x$ can be written as function of time $t$, $x = x(t)$. By the multivariate chain rule,
$$
\partial\_x f \cdot x'(t) + \partial\_t f = 0 \;.
$$
This yields an *explicit* formula for $x'(t)$, if $\partial\_x f \neq 0$ at the point of evaluation.


A similar argument works also in the multivariate case, when $\mathbf{x} \in \mathbb{R}^n$. For any time $t$, think of $F(\cdot, t)$ as an operator on $\mathbb{R}^n$, and the argument follows.


I'm interested in derivatives $\frac{d^l\mathbf{x}}{dt^l}$ of higher orders $l > 0$, in the multivariate case. Based on the above intuition, one might expect a similar formula to exist, allowing to calculate $\frac{d^l\mathbf{x}}{dt^l}$ in terms of the derivative tensors of $F$. The purpose being, to calculate high-order time derivatives at a given operator root.


**Question:**
Before setting off to derive such a formula (if exists), is there such a result in the literature? Perhaps something similar?


**Edit:**
I am aware of the Faà di Bruno's formula, and of its multivariate counterparts.


**Edit 2:**
While an expression for the first-order derivative $\frac{d\mathbf{x}}{dt}$ is commonly found in the literature when discussing the implicit function theorem as @RyanBudney notes, this question is about derivatives of *higher* orders.



Calculating derivatives of arbitrary-order at an operator's root
Although this question sounds quite innocent, a systematic treatment of higher order derivatives of implicit functions is quite involved. On a second thought, this is no surprise if you think about how complicated higher derivatives of concatenations $f\circ g$ get (see [Faà di Bruno's formula](https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula)). For higher order derivatives of implicit functions, you can have a look at [The Combinatorics of Higher Derivatives of Implicit Functions](https://arxiv.org/abs/1711.07501).


