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I thought that the meaning of $$ \frac{\partial f(x, y, z)}{\partial x} $$ is differentiation on $x$ with fixed $y$ and $z$. So $(x, y, z)$ in the numerator is just saying which variables are fixed. If I need to indicate where the derivative is evaluated, I write it in the right of a vertical bar as a subscript. But today my teacher used $(x, y, z)$ in the numerator to denote where the derivative is evaluated. So, for example, $$ \frac{\partial f(0, 0, 0)}{\partial x} $$ means $$ \frac{\partial f(x, y, z)}{\partial x} \bigg\rvert_{x=0,y=0,z=0} $$ Is that a standard convention? If so, what is the meaning of this? $$ \frac{\partial f(x, y, g(x, y))}{\partial x} $$ I have two candidates. One is a partial derivative of the composition of $f$ and $g$ where $g$ has some fixed value, and the other is the partial derivative of $f$ on $x$ evaluated at $(x, y, g(x, y))$. I think the two are not the same.

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    $\begingroup$ The same (or a closely related) ambiguity appears in the famous Euler Lagrange equations in physics also . . This problem is everywhere . . . $\endgroup$
    – Kartik
    Mar 24, 2017 at 14:09
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    $\begingroup$ Recommendation: define $\partial_i$ as the partial derivative with respect to the $i$th argument, i.e. $$ \partial_i f(x_1,x_2,\dotsc) = \lim_{h \to 0} \frac{f(x_1,x_2,\dotsc, x_i+h,\dotsc ) - f(x_1,x_2,\dotsc, x_i,\dotsc ) }{h}. $$ This avoids ambiguity, since it doesn't have anything to do with $x_i$ itself: it's just the derivative of the function $f$. The fraction notation for derivatives has similar bound variable problems to the notation for integrals, it's just less obvious. $\endgroup$
    – Chappers
    Mar 25, 2017 at 1:16
  • $\begingroup$ Last paragraph: en.wikipedia.org/wiki/Partial_derivative#Notation $\endgroup$
    – user541686
    Mar 25, 2017 at 17:35
  • $\begingroup$ I asked similar question some time ago and I got an excellent answer by DanielV. math.stackexchange.com/questions/1091097/… $\endgroup$
    – tom
    Mar 27, 2017 at 18:46

6 Answers 6

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Congratulations, you have met one of the worst ambiguities in mathematical notation!

Assume you have a function of two variables, $f \colon A \times B \to \mathbb{R}$, where $A$ and $B$ are subsets of $\mathbb{R}$. The notation $$\frac{\partial f}{\partial x}(x_0,y_0)$$ is commonly used to denote the value of the partial derivative of $f$ with respect to the first variable, evaluated at $(x_0,y_0)$. This is the cleanest use of the notation for partial derivatives.

Anyway, it sometimes happens to use some lazy piece of notation such as $$\frac{\partial f(x,g(x,y))}{\partial x}$$ to denote the partial derivative of the map $(x,y) \mapsto f(x,g(x,y))$. This is imcompatible (in general) with the interpretation of the same formula as

The derivative of $f$ with respect to the first variable, evaluated at the point $(x,g(x,y))$.

This is bad, but it seems we have to live with it. Why? Just spend a couple of minutes and think about the second interpretation. To be rigorous, we should have written $$ \frac{\partial}{\partial x} \left( f \circ \left( (x,y) \mapsto (x,g(x,y)) \right) \right) (x,y), $$ which is a true nightmare.

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    $\begingroup$ In physics text, I would immediately think $\frac{\partial}{\partial x} f(x, g(x,y))$ means $\frac{\partial f}{\partial a}(a,b)$ evaluated at $(x, g(x,y))$, and the other case would be (unambiguously) denoted by $\frac{d}{d x} f(x, g(x,y))$. An example of this distinction is the Euler-Lagrange equation which is often written written as $\frac{\partial \mathcal{L}}{\partial x} = \frac{d}{dt} \frac{\partial \mathcal{L}}{{\partial\dot{x}}}$, where $\mathcal{L}(t,x,\dot{x})$ is the Lagrangian. $\endgroup$
    – JiK
    Mar 24, 2017 at 15:16
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    $\begingroup$ @JiK In the Euler-Lagrange equation, though, both $x$ and $\dot x$ are functions of $t$. In your example above, unless $y$ is a function of $x$ it wouldn't make sense to write it as a total derivative. $\endgroup$
    – Carmeister
    Mar 24, 2017 at 18:52
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    $\begingroup$ Why not put the $f$ inside of the "lambda" instead doing it the roundabout way with a composition? (Talking about your last example) $\endgroup$ Mar 25, 2017 at 11:06
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    $\begingroup$ Can't we just say $\frac{df(g(x))}{dx}$ means to use chain rule and $\left(\frac{df}{dx}\right)(g(x))$ or $\frac{df}{dx}(g(x))$ means to plug $g$ into the derivative? $\endgroup$
    – Alex Jones
    Mar 25, 2017 at 11:10
  • $\begingroup$ The most important thing to remember is that we must not confuse the name of a function with its value at a point. In my opinion any piece of notation that goes in the opposite direction should be avoided in contemporary mathematics. $\endgroup$
    – Siminore
    Mar 25, 2017 at 13:12
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I think Siminore's answer is good. But I checked some textbooks just for curiosity.

  1. "Advanced calculus" by Folland uses the notation like $\partial_x f(0, 0)$. Of course the meaning is the partial derivative of $f$ w.r.t. $x$ evaluated at $(0, 0)$. It does not use the notation $\partial f(0, 0) / \partial x$ extensively but there is a comment that you can use the notation.
  2. "Advanced calculus" by Kaplan and "The way of analysis" by Strichartz also follow the same convention.
  3. "Real mathematical analysis" by Pugh uses the notation $\partial f(0, 0) / \partial x$ and in some place it uses $\partial f(x, g(x)) / \partial x$ to denote the partial derivative of $f$ w.r.t. $x$ evaluated at $(x, g(x))$. It doesn't mean the derivative of the composite function.
  4. I also checked "Principles of mathematical analysis" by Rudin. It looks like to avoid the notation $\partial f(a, b) / \partial x$. Instead it says "$\partial f / \partial x$ at $(a, b)$". However, actually I found only one such occurrence. There is not much use of the round notation.
  5. "Mathematical methods for physicists" by Afken uses the notation $\partial f(x, 0)/\partial t$ to denote $\partial f(x, t)/\partial t |_{t = 0}$. It uses the two notations interchangeably.

Thus, the use case of your teacher is fairly common.

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  • $\begingroup$ To top it off, $\partial f(x,y)$ is the notation for the subdifferential of $f$ at $(x, y)$... $\endgroup$
    – user541686
    Mar 25, 2017 at 7:33
  • $\begingroup$ @Mehrdad ... Or in other cases, it means the holomorphic derivative $\partial =\partial_z$. Either is unlikely to be encountered by the OP, though. $\endgroup$ Jan 15, 2019 at 1:49
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Seems like no one has mentioned that there is actually a totally unambiguous notation to deal with this problem, even though it is not in very common use:

$(\partial_1 f)(a,b,c) = \left. \dfrac{\partial f(x,y,z)}{\partial x} \right|_{(x,y,z)=(a,b,c)}$.

The "$1$" here indicates that $f$ is differentiated with respect to the first parameter. This is much better than using something like "$\partial_x$" where the $x$ is often used as a variable too and hence cannot serve well to indicate which parameter $f$ is differentiated with respect to.

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    $\begingroup$ This can be useful, but the “1” is not that less ambiguous, unfortunately. For example, Marsden & Ratiu use $D_1$ for the gradient along the first factor of a function on $X\times Y$, but then for a function on $X\times Y\times Z$ they sometimes need it as if for $X\times (Y\times Z)$ and sometimes for $(X\times Y)\times Z$. There’s no ambiguity if you’re exclusively numbering single coordinates, but that’s only useful for doing coordinate calculations, and even then you’re more likely to have a summation index than an actual number there. $\endgroup$ Jan 15, 2019 at 2:00
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    $\begingroup$ @AlexShpilkin: Well, I consider your three cartesian product expressions as different, so a function's domain can only be one of them. Whichever one it is, there would be no ambiguity. So the problem in your case is actually not the differentiation notation but rather the abuse of function notation. =) $\endgroup$
    – user21820
    Jan 16, 2019 at 5:54
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  1. Yes, $\frac{\partial f(x,y,z)}{\partial x}$ is derivative w.r.t. $x$ at fixed $y,z$.
  2. $\frac{\partial f(0,0,0)}{\partial x}$ is not standard notation. Strictly speaking, it should be zero, because $f(0,0,0)$ is a constant which does not depend on $x$. Sometimes, yes, it is used as a shorthand for $\frac{\partial f(x,y,z)}{\partial x}|_{x=y=z=0}$. But you should only do that if it is extreamly clear from the context what you mean. Generally, avoid this notation.
  3. The third expression is a derivative w.r.t. $x$, but $x$ appears two times in the numerator. Using standard derivation-rules you can compute $$\frac{\partial f(x,y,g(x,y))}{\partial x} = \frac{\partial f(x,y,z)}{\partial x}\bigg\rvert_{z=g(x,y)} + \frac{\partial f(x,y,z)}{\partial z}\bigg\rvert_{z=g(x,y)} \cdot \frac{\partial g(x,y)}{\partial x}$$
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The first interpretation is not correct, since there is no way to give $g$ some fixed value if it depends on $x$. The second interpretation is correct but worded in a confusing way, so let me clarify.

Consider the function $h(x,y)=f(x,y,g(x,y))$. Then $$ \frac{\partial f(x,y,g(x,y))}{\partial x}=\frac{\partial h(x,y)}{\partial x}. $$ The result is a function of two variables, call it $k(x,y)$. To calculate e.g. $k(x,5)$ you would compute $$ k(x,5)=\frac{d}{dx}f(x,5,g(x,5)). $$ This is a derivative in the usual sense, not a partial derivative any more. By the chain rule, we can calculate $k(x,5)$ as follows: $$ k(x,5)=f_x(x,5,g(x,5)) + f_z(x,5,g_x(x,5)). $$ There is nothing special about the number $5$, it could be anything. So replacing $5$ with $y$, we get the formula $$ \frac{\partial f(x,y,g(x,y))}{\partial x}=f_x(x,y,g(x,y)) + f_z(x,y,g_x(x,y)). $$

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I find the answer of user21820 particularly nice. But a few remarks on the historical origin of this "notational mess" might be enlightening.

Jacobi popularised the partial derivatives notation with his 1840 paper De determinantibus functionalibus, where he spends 3 pages introducing the notation, explaining a certain problem he sees with it, and suggesting a solution for this notational problem. Surprisingly, the problem Jacobi saw was not the one the OP sees. Instead, it was probably Jacobi's suggested solution which leads to the problems the OP an everyone else in this thread sees!

You may read what Jacobi he had to say here. But beware that in order to understand what he was saying (and lead any reasonable discussion about it), one needs to understand the difference between how the word function was used prior to $\approx$ 1930, and how we officially use it today. I've written about this distinction here and here.

In case you don't want to read Jacobi, here's a short rephrasing of Jacobis problem by Vladimir Arnold in his Classical Mechanics book p. 258 footnote 81:

It is necessary to use the apparatus of partial derivatives, in which even the notation is ambiguous$.^{81}$

  1. It is important to note that the quantity $\partial{u}/\partial x$ on the $x, y$-plane depends not only on the function which is taken for $x$, but also on the choice of the function $y$: in new variables $(x, z)$ the value of $\partial{u}/\partial x$ will be different. One should write $$\left.\frac{\partial u}{\partial x}\right|_{y=\text{const.}} \quad \left.\frac{\partial u}{\partial x}\right|_{z=\text{const.}}$$

In the thermodynamics literature the same problem is solved with the notation

$$\left(\frac{\partial u}{\partial x}\right)_{x,y} \quad \left(\frac{\partial u}{\partial x}\right)_{x,z}$$

but Jacobi's original suggested solution was writing

$$\frac{\partial u (x,y)}{\partial x} \quad \frac{\partial u (x,z)}{\partial x}$$

(!) Combine this with his awkward choice of the letter $f$ for what Arnold denotes with $u$ and you have guaranteed confusion. (I very much suspect that it was also Jacobi who opened the gate for the awfully popular $y=y(x)$).

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