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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 $\alpha$!

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    If you let $x=5a+3b$, you'll have to find a way to express $3a+5b$ in terms of $x$... anyway, where did you encounter this construction?2011-04-26
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    $\frac{\partial}{\partial x} \exp(x\alpha) = \alpha \exp(x\alpha)$ rather than $\alpha$.2011-04-26
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    It also depends which things in the formula are variables and which are constants. It's hard to answer your question without more context. If a and b are independent variables, $\alpha,\beta$ constants, then a change of variables $x=5a+3b$, $y=3a+5b$ would do the trick.2011-04-26
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    You can differentiate a function $f$ of several variables with respect to a vector $v\neq 0$ by the formula $\frac{\partial f}{\partial v}(u) =\lim_{h\to 0}\frac{f(u+hv)-f(u)}{h}$.2011-04-26
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    Thanks @J.M. @Henry @Jim and @david. I've edited my question a bit to clarify that $a$ and $b$ are not dependent on $\alpha$ or $\beta$. So is it ok to label $x=5a+3b$ and $y=3a+5b$ or like J.M. said, do I need to write $3a+5b$ in terms of $x$?2011-04-26
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    @Jane: In your second edit, the choice of variable names is a bit confusing -- there's no reason for the integration variables to be different, and to be the same as the free variable on the outside on one side but not on the other.2011-04-27
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    Taking the variation with respect to the operator function $x(\tau)$ is OK; just like you can change variables from $a$ and $b$ to $x=5a+3b$ and $y=3a+5b$, you can form new functions $x(\tau)=5a(\tau)+3b(\tau)$ and $y(\tau)=3a(\tau)+5b(\tau)$. The thing to keep in mind, just like for the variables, is that you need to be clear about what's being treated as varying independently. Just like for variables, you can consider either varying $x(\tau)$ and $y(\tau)$ independently, or e.g. $x(\tau)$ and $a(\tau)$. In the present case, the more natural interpretation is to take $x(\tau)$ and $y(\tau)$.2011-04-27
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    And to answer your question: No, this is not true in general; it's only true under the assumption that all the operators commute; but that has nothing to do with the question about differentiating/varying with respect to sums.2011-04-27
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    @joriki I'm not sure what you mean by that there's no need for the Integrating variables to be different...2011-04-27
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    And sorry, i'm not sure why I said operators! Must be a lack of sleep. I meant Grassmann numbers!2011-04-27
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    @Jane: You used $\tau'$ as the free variable and $\tau$ as the integration variable on the left, and $\tau'$ as the free variable and $\tau'$ as the integration variable on the right. You could use the same variable name for all four; to make it clearer, you could use a different variable name for the integration variable (on both sides) than for the free variable (on both sides); but there's no point in using different names for the two integration variables, one coinciding with the free variable -- that creates an unexpected asymmetry that causes one to wonder whether it means something.2011-04-27
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    @Jane: For Grassmann numbers, I don't see why it should come out like this despite them not commuting, but I wouldn't put it beyond them.2011-04-27
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    @joriki I used to write the same $\tau$ for all, but have been told the way I wrote it above is the correct way since when they are all tau, it looks like you're differentiating wrt the integration variable.2011-04-27
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    @joriki since $x$ is in front of $\alpha$, I don't think it should matter that they're Grassmann numbers.. Should it?2011-04-27
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    @Jane: You're right that it's preferable for the free variable to have a different name from the integration variable (in fact I wrote above that that makes it clearer); my comment referred to the fact that you did this on the left but not on the right, and that makes it asymmetric and confusing, even more so than if they were all $\tau$s.2011-04-27
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    @Jane: About the ordering, I'm not sure; I haven't used Grassman variables in quite a while; but the potential problem is not just about where $x$ is but also about where $\alpha$ is -- why do you think it should end up in front of the exponential and not behind it? I see no reason for this choice, but that might just be me.2011-04-27
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    @joriki Ahh so it should be $(\delta /\delta x(\tau')) Exp[\int_0^p d\tau (x(\tau) \alpha (\tau) - y(\tau) \beta (\tau))] = \alpha (\tau') Exp[\int_0^p d\tau (x(\tau) \alpha (\tau) - y(\tau) \beta (\tau))]$ ? (But possibly with the $\alpha$ after the integral rather than before).2011-04-27
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    And I'll think some more about the ordering - thank you.2011-04-27
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    @Jane: Yes, that's what I meant regarding the variable names. Regarding the order, I meant not so much that the $\alpha$ should be behind the exponential, but that the fact that there's no obvious reason to put it either in front or behind suggests that neither of these may be correct. But, again, I don't know that, it's just my general sense of symmetry speaking.2011-04-27

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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.

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    @Christian thanks for your reply .. I've edited my question a bit to clarify a few things. Is it ok to label $x=5a+3b$ and $y=3a+5b$ or like J.M. said, do I need to write $3a+5b$ in terms of $x$?2011-04-26
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    @Jane: The main point is that if you have $n$ free variables, then you have to change to a complete new set of $n$ free variables for partial derivatives to make sense. Introducing $x=5a+3b$ by itself doesn't define a meaning for $\partial/\partial x$. In the present case, you could introduce $x=5a+3b$ and then treat $x$ and $a$ as independent variables, or you could introduce $x=5a+3b$ and $y=3a+5b$ and treat $x$ and $y$ as independent variables; the meaning of $\partial/\partial x$ in these two cases will be different, and thus so, too, will the meaning of $\partial/\partial(5a+3b)$.2011-04-26
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    @joriki thanks. So since what I want is to find $\alpha Exp[(5a+3b)\alpha−(3a+5b)\beta]$ (or $\alpha Exp[x\alpha−y\beta]$ is fine too), the useful meaning for me of $\partial / \partial x$ is $\partial / \partial (5a+3b)$?2011-04-26
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    $a$, $b$, $\alpha$ and $\beta$ are also operators rather than just variables. Does this make any difference?2011-04-26
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    @Jane: Your question makes me think I didn't get my point across: "the useful meaning for me of $\partial / \partial x$ is $\partial / \partial (5a+3b)$" doesn't make any sense, since neither of these expressions has any meaning by itself unless you specify which other variables you're keeping fixed. In thermodynamics, where changes between various pairs or triples of variables are common, there is a usefu notation $(\partial f/\partial x)_y$ that means "the derivative of $f$ with respect to $x$, with $y$ held fixed".2011-04-26
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    You can see it in action e.g. here: http://en.wikipedia.org/wiki/Relations_between_heat_capacities . Without such a specification, the symbol $\partial/\partial(\text{whatever})$ is meaningless, independent of whether "whatever" is $x$, $5a + 3b$ or anything else. In your case, the most sensible interpretation of what's written seems to be "while holding $3a+5b$ fixed" (as has been mentioned in other comments).2011-04-26
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    That $a$ and $b$ are operators certainly could make a difference, yes (though not for the general issue of having to specify what you're keeping fixed). Do you know how differentiating with respect to one of these operators is being defined?2011-04-26
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    @joriki thanks. I tried to simplify the problem to much I think. I've edited my question again to hopefully make it clearer.2011-04-27
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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.

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    However, your answer also depends on what has been chosen as the OTHER variable...2011-04-26