General formula for the higher order derivatives of composition with exponential function

Suppose I have a function $x:\mathbb{R} \to \mathbb{R}$ and consider:
$$g(t) = e^{x(t)}$$
When I start differentiating with respect to $t$ I obtain:

My question is whether there is a reasonable expression for higher derivatives of $g$ as a function of $x$. I’m pretty sure this kind of a thing has its own name, but I cannot find anything… Thank you in advance!

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Another expression besides Faa di Brunos formula is stated as identity (3.56) in H.W. Gould’s Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule

Let $D_t$ represent differentiation with respect to $t$ and $x=x(t)$. Hence $D^n_t g(x)$ is the $n$-th derivative of $g$ with respect to $t$. The following holds true
D_t^n g(x)=\sum_{k=0}^nD_x^kg(x)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}x^{k-j}D_t^nx^j
In the special case
we have $$D_x^kg(x)=D_x^k e^x=e^x$$ and obtain

Let’s look at a small example in order to see formula (1) in action

Example: $n=2$
in accordance with OPs expression.

You can condense your findings as follows:

Write $x^{(k)}(t)=:y_k$ $(k\geq1)$. Then there is a sequence $(P_n)_{n\geq0}$ of polynomials $$P_n:=P_n(y_1,y_2,\ldots, y_{n-1})$$ such that
$${d^n\over dt^n}\bigl(e^{x(t)}\bigr)=e^{x(t)}\>P_n(y_1,\ldots, y_{n-1})\ .$$
It is easily checked that the $P_n$ satisfy the recursion
$$P_0=1,\qquad P_{n+1}(y_1,\ldots ,y_n)=y_1P_n(y_1,\ldots y_{n-1})+\sum_{k=1}^{n-1}P_{n.k}(y_1,\ldots y_{n-1})\>y_{k+1}\ ,$$
where $P_{n.k}$ denotes the partial derivative of $P_n$ with respect to the $k$th variable.

I think is this what you are looking for

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