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I still have a little problem with notation for partial derivatives.

Let $$ f(x,y) = x^2y $$

What do you think that this should equal to? $$ \frac{\partial f}{\partial x}(y,x) =\, ? $$ There are two options $2yx$ or $y^2$.

Do you think that following is the same? $$ \frac{\partial f(y,x)}{\partial x}= \,? $$ And now take substitution $g(x,y)=f(y,x)$. What is following? $$ \frac{\partial g}{\partial x}(x,y)=\,? $$ I would love to hear your opinions?


Based on DanielV comment, I need answer for things like these

$$ \frac{\partial f(y,z)}{\partial z} $$ $$ \frac{\partial f(f(y,z),z)}{\partial z} $$


I get constantly confused during physics lectures because of this :(

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    $\begingroup$ One simply just avoid writing things like $\frac{\partial f}{\partial x}(y,x)$. I can see no reason to write like that... $\endgroup$
    – mickep
    Jan 4, 2015 at 19:33
  • $\begingroup$ Yeah. Seems like just a bad notation. $\endgroup$
    – chriseur
    Jan 4, 2015 at 19:34
  • $\begingroup$ Do it in steps. 1) $\frac{df}{dx}(x,y) = 2xy$ 2) $\frac{df}{dx}(y,x) = \frac{df}{dx}(x=y,y=x) = 2yx$. $\endgroup$
    – Winther
    Jan 4, 2015 at 19:34
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    $\begingroup$ Physicists and engineers tend to conceptualize mathematics in terms of systems of variables rather than in terms of functions, so that might not work Winther. For example, if they had written: $$\frac{\partial }{\partial z} f(y, z)$$, then you can't assume that the denominator variable is a reference to the function parameter rather than a reference to the function argument. $\endgroup$
    – DanielV
    Jan 4, 2015 at 19:37
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    $\begingroup$ Alteratively, use the notation $D_1f(x,y)$: take the derivative with respect to the first variable and evaluate in $(x,y)$. $\endgroup$
    – user370967
    Sep 23, 2018 at 6:59

5 Answers 5

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Some terminology you should know:

  • Parameters are the bound variables in function definitions. So in defining $f(x, y) = x + y + z$, $x$ and $y$ are parameters, they have no meaning outside of being placemarkers defining $f$
  • Arguments are that values that are used to fill in parameters. So if I assert $\forall z~~~f(z, \pi) = f(\pi, z)$, then $\pi$ and $z$ are arguments to the function $f$ here. The scope may be global like $\pi$ or local to the expression like $z$

In common practice, when it is written

$$\frac{\partial f}{\partial x} $$

it is an abuse of notation. The idea of $x$ doesn't exist outside of defining $f$, what is indicated is the derivative of $f$ with respect to it's first argument, regardless of the name. When you see $\frac{\partial f}{\partial x} $, you should ask "where was the $x$ located when $f$ was defined?"

Physicists and engineers tend to view large problems a bit differently than mathematicians (at least those who haven't studied outside of mathematics), so things can become complicated. Real world problems tend to manifest in terms of variables like gravity, distance, time, voltage, etc, rather than functions. If you try to contort these variables and turn them into functions to conform to the calc textbooks then you are in for a world of hurt. For example, gravity could be given as a function of distance and mass, or in terms of mass and temperature where temperature is a function of distance, etc. There is no universal function that represents a variable, but variables are what must be described with the differential calculus.

So engineers and physics might actually mean

$$\frac{\partial f(y, x)}{\partial x} $$

to be interpreted as a derivative by $x$ the argument, not $x$ the parameter. You might have to ask for clarification. As mickep wisely said in the comments, it is best to avoid the reuse of variable names whenever you can. But don't forget an important point of mathematics: it isn't about presenting arguments to be memorized, it is about presenting arguments to be understood. Most ambiguous notations that occur do so because if it confuses you, then you probably aren't understanding to begin with. So before asking for clarification, first ask yourself which interpretation makes more sense in context. Then if it still isn't clear, then ask for clarification.


So the question was updated to ask about:

$$\frac{\partial f(f(y,z),z)}{\partial z}$$

presumably under the definition $f(x, y) = x^2y$. In proper mathematical conversation, $\frac{\partial}{\partial}$ is used to refer to the derivative of a function with respect to one of its parameters. If you don't use it to mean that, then the proper folk will not invite you to their putnam exam parties.

However, in practice, sometimes something like

$$Q = \frac {\partial f(f(w,z),z)}{\partial z}$$ is used in yet another abuse of notation. This one would probably mean:

$$g(w, z) = f(f(w,z),z) \quad \text{ and } \quad Q = \frac{\partial g}{\partial z}$$

In terms of variables (rather than functions), it would mean something like "the rate of change of $f(f(w,z),z)$ with respect to $z$, assuming $\frac{dw}{dz} = 0$

The important thing to watch out for with something like this is that you can't come back later and say $w = h(v,z)$, because your use of partial derivative notation assumed this dependency didn't exist. If you add a dependency like that in later, then you have to start over from scratch with:

$$g(v,z) = f(f(h(v,z),z),z) \quad \text{ and } \quad Q = \frac{\partial g}{\partial z}$$

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    $\begingroup$ Excellent! I totally dig that $\frac{\partial f}{\partial x}$ is abuse of notation. $\endgroup$
    – tom
    Jan 4, 2015 at 20:23
  • $\begingroup$ I most often see the differential operator operate on the expression and not the function. So in the case of $\partial f(f(y,z), z) \over \partial z$. I would evaluate the upper part: $f(f(y,z),z) = f(y^2 z, z) = y^4 z^3$, then differentiate this with respect to $z$ which will be $3y^4 z^2$. $\endgroup$
    – Calmarius
    Apr 1, 2017 at 19:14
  • $\begingroup$ @Calmarius It is almost a waste of time to clarify the correct evaluation procedure because half the writers aren't going to do it correctly anyway. But, one of the big differences between $\frac{\partial f(g(x))}{\partial x}$ and $\frac{{\rm d}f(g(x))}{{\rm d}x}$ is that in the first, you take the derivative of $f(x)$ wrt $x$ before applying to $g(x)$, and in the second, you apply $f$ to $g(x)$ then take the derivative wrt $x$. $\endgroup$
    – DanielV
    Apr 1, 2017 at 19:35
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    $\begingroup$ @DanielV After couple of years, I was reading a physics book yesterday and it contains some mind boggling variable change. At first, I was confused but then I recalled your answer and everything became clear. So I want to thank you again for your answer, it was really eye opening for me. $\endgroup$
    – tom
    Apr 4, 2018 at 20:08
  • $\begingroup$ There was no parameter $w$ in the last example, so I edited the example (pending review) to remove $w$ from the notation; but then to keep it an example of partial differentiation, I put in a parameter $v$ (going back to the definition, in this last example, of $w$.) I hope that this is welcome. $\endgroup$ Sep 23, 2018 at 4:51
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By definition, $\frac{\partial f}{\partial x}$ is a function and

$$\frac{\partial f}{\partial x} (y, x)$$

means that you plug in $(y, x)$ to evaluate your function. Thus it should be $2xy$.

Think it this way: What is

$$\frac{\partial f}{\partial x} (1, 0)?$$

Obviously you do not first plug in $x=1$ and $y=0$ and then differentiate wrt $x$.......

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Actually,

$$\frac{\partial f}{\partial x}(y,x) $$ is the derivative over $x$ evaluating at the point $(y,x).$

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  • $\begingroup$ Correct. And since it produces a derivative for all $(x,y)$ where it's defined, it's a function in $\mathbb R^2$. $\endgroup$
    – Jon
    Jan 4, 2015 at 19:37
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    $\begingroup$ Whether it is a function or not is ambiguous. Unless you are actually going to use lambda notation for function definitions (and you don't want to see that in the context of differential calculus), then you have to settle for guessing. $\endgroup$
    – DanielV
    Jan 4, 2015 at 20:11
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If $f$ is defined by $f(x,y)$, the operation $\frac{\partial f}{\partial x}$ means that you differentiate $f$ with respect to $x$. With your example this will become $2xy$. Now, $\frac{\partial f}{\partial x}(y,x)$ means, that you FIRST differentiate with respect to $x$ (which was $2xy$) and AFTERWARDS plug in $(x,y)=(y,x)$. Therefore we obtain $\frac{\partial f}{\partial x}(y,x)=2yx(=2xy)$.

Now, if $g(x,y)=f(y,x)$, then $g(x,y)=y^2x$. Again, $\frac{\partial g}{\partial x}(x,y)$ means that you have to differentiate $g$ with respect to $x$ and afterwards plug in $(x,y)=(x,y)$ (so this won't change anything). Thus, we obtain $\frac{\partial g}{\partial x}(x,y)=y^2$.

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    $\begingroup$ This is not correct. The notation $\frac {\partial f}{\partial x}$ means to differentiate $f$ with respect to it's first parameter. The use of $x$ is an abuse of notation, as $x$ has no meaning outside of the declaration of the definition of $f$. $\endgroup$
    – DanielV
    Jan 4, 2015 at 20:12
  • $\begingroup$ I refered to the op's definition of $f$, so I implicitly assumed that $f$ is known to be a function of the two variables $f(x,y)$. $\endgroup$
    – sranthrop
    Jan 4, 2015 at 21:20
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Strictly going by the rule that in $\frac{\partial f}{\partial *}(*,\cdots,*)$, you always perform the derivative operation first and then worry about evaluation, the answer must be $2xy$. This rule must be followed, otherwise, expressions like $\frac{\partial f}{\partial y}(17)$ don't make any sense. The Leibnitzian notation is an unfortunate one to begin with and its extension to partial derivatives is bordering on nonsense. The Eulerian notation really shows its virtues in these cases. For the example above, we would write $D_1 f(y,x)$, which is completely unambiguous.

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