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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.

Mutantoe
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LuianChu
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    The same (or a closely related) ambiguity appears in the famous Euler Lagrange equations in physics also . . This problem is everywhere . . . – Kartik Mar 24 '17 at 14:09
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    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. – Chappers Mar 25 '17 at 01:16
  • Last paragraph: https://en.wikipedia.org/wiki/Partial_derivative#Notation – user541686 Mar 25 '17 at 17:35
  • I asked similar question some time ago and I got an excellent answer by DanielV. http://math.stackexchange.com/questions/1091097/partial-derivative-notation – tom Mar 27 '17 at 18:46

6 Answers6

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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.

Siminore
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    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. – JiK Mar 24 '17 at 15:16
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    @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. – Carmeister Mar 24 '17 at 18:52
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    Why not put the $f$ inside of the "lambda" instead doing it the roundabout way with a composition? (Talking about your last example) – Luka Horvat Mar 25 '17 at 11:06
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    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? – Alex Jones Mar 25 '17 at 11:10
  • 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. – Siminore Mar 25 '17 at 13:12
  • The last one doesn't look right. You just wrote $\frac{\partial}{\partial x} r(x, y)$ for some complicated $r$, which is still evaluating the partial derivative at the point $(x, y)$. If you meant to get the partial derivative of $r$ then I think you should've omitted the final $(x, y)$? – user541686 Mar 25 '17 at 17:31
  • @Siminore, your rigorous suggestion is not rigoros enough: when $f:A\times B \to\mathbb{R}$, then the notation $\frac{\partial{f}}{\partial x}$ is a bad idea, since $f$ doesn't care how we "name the input variables". It would only be rigorous if we declare by convention, that the $x$ in the notation always denotes derivative with respect to the first component of the product $A\times B$. But that seems arbitrary. Better write $\partial_1 f$ as suggested by Chappers in the comments. – Michael Bächtold May 17 '18 at 19:03
  • @MichaelBächtold Well, if I denote $f \colon (x,y) \mapsto \ldots$, the name of the variables is clearly fixed. Anyway, I agree that using a numberl like $\partial_1 f$ is a good choice. – Siminore May 18 '18 at 07:30
  • @Siminore in the usual (or at least my) understanding of the map notation $(x,y)\mapsto f(x,y)$ the variables xy are bound. So for example I’d find it correct to write $f= ((x,y)\mapsto f(x,y))=((y,x)\mapsto f(y,x))$, which means it would be very confusing to write $\partial_x f$. But to be fair, my interpretation of the \mapsto is probably more established in the lambda calculus circles, and not so much among average mathematicians. – Michael Bächtold May 18 '18 at 08:47
  • @MichaelBächtold In my understanding, $f \colon (# 1,# 2) \mapsto f(# 1,# 2)$, so we could write $\partial_{# 1}f$. But I agree that writing $\partial_x f$ is essentially meaningless, while it would meaningful to write $\partial_x ((x,y) \mapsto f(x,y))$. – Siminore May 18 '18 at 11:26
  • I'm not completely sure how to read that notation. Would you consider the map $(#1,#2)\mapsto \sin(#1)+#2$ to be the same as the map $(#2,#1)\mapsto \sin(#2)+#1$? – Michael Bächtold May 18 '18 at 11:31
  • @MichaelBächtold Of course, just think of $# 1={#\ }1$. It is only a symbol. What I am trying to say is this: if you consider a generic function $f \colon X \times Y \to Z$, then it is unclear what $\partial_y f$ means. But if you write $\partial\beth ((\aleph,\beth)\mapsto f(\aleph,\beth))$, then this is formally correct. The notation $\partial_j f$ for a derivative of $f \colon X_1 \times \cdots \times X_n \to Z$ is always clear. – Siminore May 18 '18 at 11:33
  • But that suggestion contradicts a standard convention of bound variables: alpha conversion. If the object $(x,y)\mapsto f(x,y)$ is indistinguishable from $(y,x)\mapsto f(y,x)$ then I should be allowed to substitute one for the other in any context. So we would have that $\partial_x((x,y)\mapsto f(x,y))$ is identical to $\partial_x((y,x)\mapsto f(y,x))$, wouldn't we? – Michael Bächtold May 18 '18 at 11:42
  • By the way, the confusing notation $\frac{\partial f}{\partial x}=\frac{f(x,y)}{\partial x}$ has an amusing history: Jacobi introduced it with the intention of being more precise. He saw a problem with the notation of partial derivatives which is seldom discussed in modern calculus courses, but it's mentioned in several places here on math.SE. Unfortunately, Jacobis cure was worse that the disease. – Michael Bächtold May 18 '18 at 12:03
  • @MichaelBächtold No, the first is $\partial_1$, the second is $\partial_2$. You are actually defining the same map: why should $x$ come before $y$? – Siminore May 18 '18 at 13:02
  • "Why should $x$ come before $y$?" In the notation $(x,y)\mapsto f(x,y)$ I'm allowed to rename the bound variables at will without changing the meaning of the expression. In particular I'm allowed to swap $y$ and $x$. I believe we agree on that. Would you also agree that if two expressions denote the same thing (like $f$ and $(a,b)\mapsto f(a,b)$) then I should be allowed to substitute one expression for the other inside any larger expression without changing the meaning of the larger expression? So for example $\partial_1 f$ should have the same meaning as $\partial_1 ((a,b)\mapsto f(a,b))$. – Michael Bächtold May 18 '18 at 13:26
  • If you agree with this, I don't see how you can also accept $\partial_x ((x,y)\mapsto f(x,y))$ as a formally correct notation, since then $\partial_x ((a,b)\mapsto f(a,b))$ should mean the same, as should $\partial_x ((y,x)\mapsto f(y,x))$, by this rule of substituting equals for equals. – Michael Bächtold May 18 '18 at 13:26
  • In your first example, there is no $x$ variable in the definition of $f$... Hence the $x$-derivative is always null. In the second, I add a small quantity to $x$, so that I am moving the second argument of $f$... Of course in this piece of notation you cannot change variables as you like. Once you have chosen the names inside the brackets, you must keep them untile you start with a different computation. I repeat: $(x,y)\mapsto f(x,y)$ is a local procedure that freezes variables untile you decide that the procedure is finished. This is very different from $f \colon X \to Y$. – Siminore May 18 '18 at 13:32
  • Let us continue this discussion in chat. – Michael Bächtold May 18 '18 at 16:45
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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.

leeto
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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.

user21820
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    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. – Alex Shpilkin Jan 15 '19 at 02:00
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    @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. =) – user21820 Jan 16 '19 at 05:54
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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 $$\frac{\partial f(x,y,g(x,y))}{\partial x} = \frac{\partial f(x,y,z)}{\partial x}\bigg\rvert_{z=g(x,y)} + \frac{\partial f(x,y,z)}{\partial z}\bigg\rvert_{z=g(x,y)} \cdot \frac{\partial g(x,y)}{\partial x}$$
Simon
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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 $$ \frac{\partial f(x,y,g(x,y))}{\partial x}=f_x(x,y,g(x,y)) + f_z(x,y,g_x(x,y)). $$

pre-kidney
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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 $y=y(x)$).

Michael Bächtold
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