# 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:
\begin{align}
g’&=e^xx’\\
g”&=e^x((x’)^2+x”)\\
g”’&=e^x((x’)^3+3x’x”+x”’)\\
&…
\end{align}

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!

#### Solutions Collecting From Web of "General formula for the higher order derivatives of composition with exponential function"

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
\begin{align*}
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
\end{align*}
In the special case
\begin{align*}
g(x(t))=e^{x(t)}
\end{align*}
we have $$D_x^kg(x)=D_x^k e^x=e^x$$ and obtain
\begin{align*}
D_t^ne^x=e^x\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}x^{k-j}D_t^nx^j\tag{1}
\end{align*}

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

Example: $n=2$
\begin{align*}
D_t^2e^x&=e^x\sum_{k=0}^2\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}x^{k-j}D_t^2x^j\\
&=e^x\left(\frac{(-1)^1}{1!}\left(-\binom{1}{1}x^0D_t^2x\right)
+\frac{(-1)^2}{2!}\left(-\binom{2}{1}xD_t^2x+\binom{2}{2}x^0D_t^2x^2\right)\right)\\
&=e^x\left(D_t^2x+\frac{1}{2}\left(-2xD_t^2x+D_t^2x^2\right)\right)\\
&=e^x\left(x^{\prime\prime}+\frac{1}{2}\left(-2xx^{\prime\prime}+D_t\left(2xx^{\prime}\right)\right)\right)\\
&=e^x\left(x^{\prime\prime}+\frac{1}{2}\left(-2xx^{\prime\prime}+2\left(x^{\prime}\right)^2+2xx^{\prime\prime}\right)\right)\\
&=e^x\left(x^{\prime\prime}+\left(x^{\prime}\right)^2\right)
\end{align*}
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