Can you differentiate with respect to a sum? Is the following true?
\begin{equation}
\frac{\partial}{\partial (5a+3b)} \operatorname{Exp}[(5a+3b)\alpha - (3a+5b)\beta] = \alpha \operatorname{Exp}[(5a+3b)\alpha - (3a+5b)\beta] .
\end{equation}
Can you differentiate with respect to a sum like this? (Or perhaps equivalently substitute $x=5a+3b$?). Note that $a$ and $b$ are not dependent on $\alpha$ or $\beta$, but all of the terms, $a$, $b$, $\alpha$ and $\beta$ are all dependent on a further variable, $\tau$. The equation above is a simplified version of what I actually want to find, which is a functional derivative.
Edit2: So my non-simplified question is whether this is true:
\begin{equation}
\frac{\delta}{\delta x(\tau')} \operatorname{Exp}[\int_0^p d\tau (x (\tau) \alpha (\tau) - y (\tau)\beta (\tau))] = \alpha (\tau') \operatorname{Exp}[\int_0^p d\tau' (x (\tau')\alpha(\tau') - y (\tau')\beta (\tau'))] .
\end{equation}
where $x (\tau)=5a (\tau)+3b (\tau)$, $y(\tau)=3a (\tau)+5b (\tau)$ and each of $\alpha$, $\beta$, $a$ and $b$ (and therefore $x$ and $y$) are operators. - Does it matter that the functional derivative is with respect to a sum?
Thanks.
Edit1: Thanks @henry for spotting my typo. - The right hand side of the equation is $\alpha \operatorname{Exp}[(5a+3b)\alpha - (3a+5b)\beta]$ not just $\alpha$!
 A: If you are in an $n$-dimensional environment (here $n=2$) you only can talk about partial derivatives like ${\partial\over \partial x}$, ${\partial\over \partial y}$ or ${\partial\over \partial x_k}$ after you have chosen $n$ coordinate functions $x_k$ (resp. $x$ and $y$ in the case $n=2$). Maybe in your case the two intended coordinate functions are $x:=5a+3b$ and $y:=3a+5b$. Anyway, before one has agreed to ${\it both}$ $x$ and $y$  it is ${\it forbidden}$ to talk about ${\partial \over \partial x}$ or ${\partial \over \partial y}$. So let's assume we have chosen $x$ and $y$ as mentioned. Then the left side of your expression amounts to
${\partial\over \partial x}\exp(\alpha x+\beta y)$, and this by the rules of calculus computes to $\alpha \exp(\alpha x+\beta y)$, which is a function of $x$ and $y$ and not the constant $\alpha$ proposed in your question.
A: Yes, assuming the conditions of the implicit function theorem are satisfied. The easiest way to show this is to make the change of variables so that the derivative is expressed in terms of one of the new variables. If an implicit function exists, then it is always possible to make such a change of variables.
