I just ran across the following expression and would like to know if anyone can identify it:
$\sum\limits_{n=1}^\infty \frac{(-1)^n}{n!}\frac{d^n f(x)}{dx^n}\frac{d^n g(y)}{dy^n}$.
It almost looks like a shift operator,
$T^tf(x)=f(x+t)=e^{t\frac{d}{dx}}f(x)$,
but is a bit more complicated.
This is considerably more involved than the shift. The expression (with $-1$ replaced by $t$) $$ F(t) = \sum_{n=0}^{\infty} \frac{t^n}{n!}\frac{d^nf}{dx^n}\frac{d^ng}{dy^n} $$ solves the PDE $$ \dot{F} = \frac{\partial^2}{\partial x\partial y} F , \quad\quad F(0)=f(x)g(y) $$ (as already observed by Gerard, in slightly different form). You are interested in $F(-1)-f(x)g(y)$.
The operator on the RHS is self-adjoint (on $L^2$), but unbounded above and below, so we can solve this by the usual Hilbert space methods, but the evolution operator will not be bounded and thus will have a domain.
We can change variables to $u=x+y$, $v=x-y$ to obtain $$ \dot{F} = F_{uu} - F_{vv}, \quad\quad F(0)=f\left( \frac{u+v}{2}\right) g\left(\frac{u-v}{2}\right) . $$ Since these two one-dimensional Laplacians $\Delta_u$, $\Delta_v$ commute, we obtain $F(t)$ from $F(0)$ by applying two one-dimensional heat evolutions, but one of them (the one associated with the $v$ variable) goes the "wrong" way, into the past, so this only makes sense for initial conditions in the domain of this evolution; compactly supported Fourier transform in the $v$ variable would be sufficient.
One has $$ \sum\limits_{n=1}^\infty \frac{(-1)^n}{n!}\frac{d^n f(x)}{dx^n}\frac{d^n g(y)}{dy^n}=(exp(-(\frac{d^2}{dxdy}))[f(x)g(y)]-f(x)g(y) $$
the one-parameter corresponding group is $e^{t\frac{d^2}{dxdy}}=\Phi(t)$ acting on bivariate series as the sum you gave is
$$ \Phi(-1)[f(x)g(y)]-f(x)g(y) $$ $\Phi(t)$ always converges with $t$ formal of $t$ real/complex on entire series. To integrate it, you can begin considering the functions $f(x)e^{y}\mapsto f(x+t)e^{y};\ e^xg(y)\mapsto e^{x}g(x+t)$. Hope this helps.
Note that, using eigenfunctions of the derivative to conjugate $\Phi(t)$, one has a differential expression of the "time-frequency" transform. Indeed $$ e^{-ay}\Phi(t)[f(x)e^{ay}]=f(x+at) $$ which acts on entire functions and corresponds to a dilation of the time by $a$ and a shift by $x$.
