Derivatives of $exp(f(x,y))$

Question

I tried unsuccessfully to find this information in other posts, so I'll ask it here:

The $n$-th derivative of $e^{f(x)}$ can be given in terms of the complete Bell polynomials:

$$ \begin{align} \frac{d^n}{dx^n}e^{f(x)} &= e^{f(x)}B_n(f^{(1)}(x),...,f^{(n)}(x))\\ &= e^{f(x)}B_n^{(1)}(f) \end{align} $$

where $f^{(n)}(x) = \frac{d^n}{dx^n}f(x)$ and $B_n^{(1)}(f)$ indicates that we are considering the derivatives of $f$ with respect to its 1st argument.

I need the two-variable version of this formula, i.e.

$$ \begin{align} \frac{d^m}{dy^m}\frac{d^n}{dx^n}e^{f(x,y)} &= \frac{d^m}{dy^m}e^{f(x,y)}B_n(f^{(1,0)}(x,y),...,f^{(n,0)}(x,y))\\ &=\frac{d^m}{dy^m}e^{f(x,y)} B_n^{(1)}(f)\\ &=\sum_{k=0}^m \binom{m}{k}\frac{\partial^k e^{f(x,y)}}{\partial y^k}\frac{\partial^{m-k} B_n^{(1)}(f)}{\partial y^{m-k}}\\ &=e^{f(x,y)}\sum_{k=0}^m \binom{m}{k}B_k^{(2)}(f)\frac{\partial^{m-k} B_n^{(1)}(f)}{\partial y^{m-k}} \end{align} $$

So all I need to compute are the derivatives of $B_n(f^{(1,0)}(x,y),...,f^{(n,0)}(x,y))$ with respect to $y$.

We can consider the derivatives $f^{(j,0)}(x,y)$ as independent functions of $y$, so all in all I need a way to compute the terms:

$$ \frac{\partial^k}{\partial y^k}B_n(h_1(y),...,h_n(y)) $$

Answer 1

We obtain

\begin{align*} \color{blue}{\frac{d^m}{dy^m}}&\color{blue}{B_n\left(x_1(y),x_2(y),\ldots,x_n(y)\right)}\\ &=\frac{d^m}{dy^m}\sum_{k=1}^nB_{n,k}\left(x_1(y),x_2(y),\ldots,x_{n-k+1}(y)\right)\\ &=\frac{d^m}{dy^m}\sum_{k=1}^n\sum_{{j_l\geq 0,1\leq l\leq n-k+1}\atop{{\sum_{l=1}^{n-k+1}j_l=k}\atop{\sum_{l=1}^{n-k+1}lj_l=n}}}n!\prod_{l=1}^{n-k+1} \frac{\left(x_l(y)\right)^{j_l}}{j_l!l!}\tag{1}\\ &=n!\sum_{k=1}^n\sum_{{j_l\geq 0,1\leq l\leq n-k+1}\atop{{\sum_{l=1}^{n-k+1}j_l=k}\atop{\sum_{l=1}^{n-k+1}lj_l=n}}} \left(\prod_{l=1}^{n-k+1}\frac{1}{j_l!l!}\right) \frac{d^m}{dy^m}\left(\prod_{l=1}^{n-k+1}\left(x_l(y)\right)^{j_l}\right)\\ &\,\,\color{blue}{=n!\sum_{k=1}^n\sum_{{j_l\geq 0,1\leq l\leq n-k+1}\atop{{\sum_{l=1}^{n-k+1}j_l=k}\atop{\sum_{l=1}^{n-k+1}lj_l=n}}} \left(\prod_{l=1}^{n-k+1}\frac{1}{j_l!l!}\right)}\\ &\quad\qquad\color{blue}{\times\sum_{{j_t\geq 0,1\leq t\leq n-k+1}\atop{\sum_{t=1}^{n-k+1}q_t=m}} \binom{m}{q_1,q_2,\ldots,q_{n-k+1}}\prod_{t=1}^{n-k+1}\frac{d^{q_t}}{dy^{q_t}}(x_t(y))^{j_t}}\tag{2} \end{align*} which is admittedly not handy, but could be used for further analysis.

Comment:

• In (1) we use a representation of the complete exponential Bell polynomial.

• In (2) we apply the general Leibniz rule.

Answer 2

Letting
\begin{align*} \frac{\partial^{m+n}}{\partial x^m \partial y^n}e^{f(x,y)} = e^{f(x,y)}T_{m,n}\qquad\qquad m\geq 0,n\geq 1\tag{1} \end{align*} we prove OPs recurrence relation

\begin{align*} T_{m,n} &= \sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}T_{r,s}f^{(m-r,n-s)}(x,y)\qquad m\geq 0,n\geq1\tag{2} \end{align*}

We obtain from (1) for $m\geq 0, n\geq 1$:

\begin{align*} \color{blue}{T_{m,n}}&=e^{-f(x,y)}\frac{\partial^{m+n}}{\partial x^m\partial y^n}e^{f(x,y)}\\ &=e^{-f(x,y)}\frac{\partial^{m+n-1}}{\partial x^m\partial y^{n-1}}\left(\frac{\partial}{\partial y}e^{f(x,y)}\right)\\ &=e^{-f(x,y)}\frac{\partial^{m}}{\partial x^m}\frac{\partial^{n-1}}{\partial y^{n-1}}\left(f^{(0,1)}(x,y)e^{f(x,y)}\right)\\ &=e^{-f(x,y)}\frac{\partial^{m}}{\partial x^m}\sum_{s=0}^{n-1}\binom{n-1}{s}f^{(0,n-s)}(x,y)\frac{\partial s}{\partial y^s}e^{f(x,y)}\tag{3}\\ &=e^{-f(x,y)}\sum_{s=0}^{n-1}\binom{n-1}{s}\frac{\partial^{m}}{\partial x^m}\left(f^{(0,n-s)}(x,y)\frac{\partial ^s}{\partial y^s}e^{f(x,y)}\right)\\ &=e^{-f(x,y)}\sum_{s=0}^{n-1}\binom{n-1}{s}\sum_{r=0}^m\binom{m}{r}f^{(m-r,n-s)}(x,y)\frac{\partial ^r}{\partial x^r}\left(\frac{\partial ^s}{\partial y^s}e^{f(x,y)}\right)\tag{4}\\ &=\sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}f^{(m-r,n-s)}(x,y)\left(e^{-f(x,y)}\frac{\partial ^{r+s}}{\partial x^r\partial y^s}e^{f(x,y)}\right)\\ &\,\,\color{blue}{=\sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}f^{(m-r,n-s)}(x,y)T_{r,s}} \end{align*}

and the claim (2) follows.

Comment:

• In (3) and (4) we apply the general Leibniz rule

Answer 3

The answer by Markus is definitely appreciated, but I may have found a simpler way.

We expect the solution to be in the form

$$ \frac{\partial^{m+n}}{\partial x^m \partial y^n}e^{f(x,y)} = e^{f(x,y)}T_{mn}, $$

where $T_{mn} = \sum_{p,q}C^{mn}_{pq}f^{(p,q)}(x,y)$ is a polynomial in the partial derivatives of $f(x,y)$. I have worked out a recursive definition of $T_{mn}$:

$$ \begin{align} T_{m0} &= B_m(f^{(1,0)}(x,y),\dots,f^{(m,0)}(x,y))\\ T_{mn} &= \sum_{r=0}^m\sum_{s=0}^{n-1}\binom{m}{r}\binom{n-1}{s}T_{rs}f^{(m-r,n-s)}(x,y)\qquad n\geq1 \end{align} $$

The only issue is that I don't have an actual proof, I found this result by generating a lot of examples with Mathematica and by staring at them long enough.