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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? –  J. M. Apr 26 '11 at 16:52
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$\frac{\partial}{\partial x} \exp(x\alpha) = \alpha \exp(x\alpha)$ rather than $\alpha$. –  Henry Apr 26 '11 at 17:00
    
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. –  Grumpy Parsnip Apr 26 '11 at 17:08
    
You can differentiate a function $f$ of several variables with respect to a vector $v\neq 0$ by the formula $\frac{\partial f}{\partial v}(u) =\lim_{h\to 0}\frac{f(u+hv)-f(u)}{h}$. –  Davide Giraudo Apr 26 '11 at 17:51
    
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 $x$? –  Jane Apr 26 '11 at 19:42

2 Answers 2

up vote 3 down vote accepted

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 $x$? –  Jane Apr 26 '11 at 19:47
    
@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 $\partial/\partial(5a+3b)$. –  joriki Apr 26 '11 at 20:04
    
@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 $\partial / \partial (5a+3b)$? –  Jane Apr 26 '11 at 20:11
    
$a$, $b$, $\alpha$ and $\beta$ are also operators rather than just variables. Does this make any difference? –  Jane Apr 26 '11 at 20:13
    
@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". –  joriki Apr 26 '11 at 20:53

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... –  GEdgar Apr 26 '11 at 23:36

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