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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.

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  • $\begingroup$ What do you mean by a "root"? Are you solving for $x$ as a function of $t$ in the level-set $\{(x,y) : f(x,t)=0\}$? If so, this is a common topic in muilti-variable calculus, when they discuss the implicit function theorem. $\endgroup$ Commented Mar 26, 2022 at 21:11
  • $\begingroup$ Yes @RyanBudney, I refer to points $(\mathbf{x}, t)$ on which $f$ vanish as roots. I'm solving for $\mathbf{x}$ in terms of $t$. In discussions on the implicit function theorem, I only found references containing an expression for the first-order derivative $\frac{d\mathbf{x}}{dt}$, while I'm after high-order ones $\frac{d^l\mathbf{x}}{dt^l}$. Are you aware of any discussions giving (some kind of) formula for derivatives of order $l > 1$? $\endgroup$
    – Shlomi A
    Commented Mar 27, 2022 at 5:19
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    $\begingroup$ Have you tried just differentiating both sides repeatedly in t, using the chain rule? That seems to work. $\endgroup$
    – Ben McKay
    Commented Mar 27, 2022 at 8:49
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    $\begingroup$ Thanks @ben-mckay. Indeed, while it seems to work, I wondered whether there already is such a result out there... $\endgroup$
    – Shlomi A
    Commented Mar 27, 2022 at 14:27

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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). For higher order derivatives of implicit functions, you can have a look at The Combinatorics of Higher Derivatives of Implicit Functions.

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  • $\begingroup$ Thank you @Dirk! That's a good point... Zemel's paper to which you refer seems to solve the univariate case of this problem. $\endgroup$
    – Shlomi A
    Commented Mar 27, 2022 at 14:43

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