# Calculus of variations question from Darcogona

I asked the question in the next forum, the 4th post, hopefully someone can help me with this here or there: https://nrich.maths.org/discus/messages/7601/151442.html?1310911861

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Please post the question here in full instead of merely linking to it. –  Ｊ. Ｍ. Jul 17 '11 at 14:31

Yes, this is what happens when authors mix partial derivatives and variable changes indiscriminately without making any effort to indicate what's being varied and what's being held fixed in a given context.

$f$ is a function of three variables, and they can all be varied independently, so there are three partial derivatives of $f$ with respect to its three arguments, the other two being held fixed. $\xi$ and $\eta$ are functions of two variables, so they each have two partial derivatives with respect to their two arguments, the other being held fixed. So far, so clear.

The second argument of both $\xi$ and $\eta$ is $\epsilon$, but the first arguments are different, so the partial derivatives of $\xi$ and $\eta$ with respect to $\epsilon$ are different derivatives in different coordinate systems. (By this I mean not just that obviously the derivatives of two different functions are different, but that two different differential operators are being applied to these two functions, both denoted by the same notational device (a subscript $\epsilon$)). When the author then writes the integrand as $g(\epsilon)$ and differentiates it with respect to $\epsilon$ by adding a prime, it's not immediately obvious which of these two derivatives is intended. However, the "change of variables $y=\xi(x,\epsilon)$" suggests that $y$ and $\epsilon$ are now to be regarded as the independent variables, and indeed the calculation of $g'$ makes sense if you interpret it that way. For your calculation, you interpreted it not just the other way, but both ways -- your expression contains both $\partial\xi/\partial\epsilon$ and $\eta_\epsilon$ -- but since these arise from applying two different differential operators, at most one of them could be the intended result of the differentiation with respect to $\epsilon$. You'll get the right result from your expression if you set $\partial\xi/\partial\epsilon$ to zero, which indeed it is if $y$ and $\epsilon$ are the independent variables, since $\xi=y$.

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This mistake is made surprisingly often. I often see people define a coordinate function, say $r$, and proceed to take the $\partial/\partial r$ derivative without ever bothering to define the rest of the coordinate system... I was at a conference last year and heard someone call this "The Second Fundamental Mistake of Multivariable Calculus". I didn't inquire to what the first is. –  Willie Wong Jul 21 '11 at 15:08