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

Question

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!

Answer 1

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.

Answer 2

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.

Answer 3

I think is this what you are looking for

http://www.wolframalpha.com/input/?i=d%5En%2Fdt%5En(e%5E(x(t)))