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There are at least two substantially different meanings to $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$. The $\partial x$ could mean "with respect to $x$ the independent variable," or it could mean "with respect to the $x$ the first parameter of $f$." I think this can be understood in light of a net income calculation. Suppose $x$ is an individual's taxable gross income, $y$ is her non-taxable gross income (gifts received, etc.), $z$ is her income tax, and $f$ is her net income, all over the same year. Since net income depends on taxable gross income, non-taxable gross income, and income tax, as given by $f = x + y - z$, and income tax depends on taxable gross income, as given by $z = .15x$ (using a single 15% tax bracket for simplicity), we can write the overall equation as $f(x,\ y,\ z(x)) = x + y - z(x)$ where $z(x) = .15x$, and then consider the meaning of $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$.

If we interpret $\partial x$ to mean "with respect to $x$ the independent variable," then $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$ represents the change in net income relative to a reported change in taxable gross income, whereas if we interpret $\partial x$ to mean "with respect to the $x$ the first parameter of $f$," then $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$ represents the change in net income relative to an unreported change in taxable gross income.

I have given, just as I have learned, a binary explanation of this difference. The $\partial x$ refers to either an independent variable or a parameter of $f$. My question is whether it is also acceptable for it to refer to something in between. Let's assign a new color to the contents of each nested layer of a function's parentheses, so that the above example becomes $f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{)}})$. This disambiguates things by allowing us to refer to the change in net income relative to a reported change in taxable gross income with $\frac{\partial}{\partial \color{tan}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{)}})$ and the change in net income relative to an unreported change in taxable gross income with $\frac{\partial}{\partial \color{blue}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{)}})$. I think colors are less misleading than subscripts in this case, because $\color{tan}{\textrm{x}}$ and $\color{blue}{\textrm{x}}$ are the same algebraic entity; it's just that when the calculus eats an algebraic expression and spits out a new one, it sometimes chews up the two $x$'s a bit differently.

With this setup, the question can be asked quite succinctly; can $\frac{\partial}{\partial x}f(x,\ y,\ z(x),\ a(x,\ z(x)), b(x,\ z(x))$ also mean $\frac{\partial}{\partial \color{orange}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{), a(}}\color{orange}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{orange}{\textrm{)}}\color{blue}{\textrm{), b(}}\color{orange}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{orange}{\textrm{)}}\color{blue}{\textrm{)}}$ and/or $\frac{\partial}{\partial \color{lime}{\textrm{x}}}f(\color{blue}{\textrm{x, y, z(}}\color{tan}{\textrm{x}}\color{blue}{\textrm{), a(}}\color{lime}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{lime}{\textrm{)}}\color{blue}{\textrm{), b(}}\color{red}{\textrm{x, z(}}\color{tan}{\textrm{x}}\color{red}{\textrm{)}}\color{blue}{\textrm{)}}$, and/or have some other meaning drawn via a similar color hierarchy, or is a partial derivative unable to be taken with respect to orange or green (or red) $x$, since they are neither independent variables nor parameters of $f$?

user10478
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  • Okie, I changed it. – user10478 Apr 23 '19 at 04:08
  • What do you mean if a partial derivative can be taken? I mean, you can take partial derivatives as you wish, as long as the functions are differentiable. Are you really asking what is the proper notation to use when one wants to convey each of the different meanings? – Jackozee Hakkiuz Apr 23 '19 at 04:33
  • I'm not necessarily expecting alternative notation. It's more "if I stumble across the above partial derivative notation, how many and what ambiguous meanings might it possibly have?" – user10478 Apr 23 '19 at 04:42
  • You can give the notation as many meanings as you want. My view is that only one is correct, though. – Jackozee Hakkiuz Apr 23 '19 at 06:12
  • Perhaps you have some unstated conditions in mind when you say "you can take partial derivatives as you wish" and "you can give the notation as many meanings as you want." For example, I assume $\frac{\partial}{\partial x}f(x,\ y,\ z(x))$ in my income case CANNOT mean "the change in net income relative to a fake reported change in taxable gross income." In other words, it can't mean $\frac{f(x,\ y,\ z(x + h))}{h}$; a nudge must propagate outward to all encapsulating functions that also explicitly contain an $x$. Is that correct? – user10478 Apr 23 '19 at 18:02
  • Programming analogies also lead me to tentatively suspect that my orange example is invalid, since although the "call stack heights" of the two orange $x$'s are identical, their "scopes" are detached from one another. – user10478 Apr 23 '19 at 18:03
  • Above should say $\frac{f(x,\ y,\ z(x + h)) - f(x,\ y,\ z(x))}{h}$. – user10478 Apr 25 '19 at 17:01

3 Answers3

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There are at least two substantially different meanings to $\frac{\partial}{\partial x}f(x, y, z(x))$.

This should not be the case. Take two differentiable functions $f:\mathbb{R}^{3}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$

The notation $\frac{\partial}{\partial x}f(x,y,g(x))$ should be understood as differentiating the function $(x,y)\mapsto f(x,y,g(x))$ with respect to every ocurrence of $x$. I.e., by the multivariable chainrule:

$$\frac{\partial}{\partial x}f(x,y,g(x)) = \left.\frac{\partial}{\partial x} f(x,y,z)\right|_{z=g(x)} + \left.\frac{\partial}{\partial z}f(x,y,z)\right|_{z=g(x)}\cdot g'(x)$$

(note: it is understood that the differentiation is performed first, and the evaluation $z=g(x)$ later). If you want to differentiate $f$ with respect to its first argument, while keeping the other ones fixed, then write $$\frac{\partial}{\partial x}f(x,y,z)$$ and if you want to compose $g$ with the third argument of $f$, and then differentiate this composition with respect to the argument of $g$, just write $\frac{\partial}{\partial a}f(x,y,g(a))$, which by the chainrule equals $$\frac{\partial}{\partial a}f(x,y,g(a))=\left.\frac{\partial}{\partial z}f(x,y,z)\right|_{z=g(a)} \cdot g'(a),$$ and then if you want to evaluate this expression at $a=x$, write $$\left.\frac{\partial}{\partial a}f(x,y,g(a))\right|_{a=x} =\left.\frac{\partial}{\partial z}f(x,y,z)\right|_{z=g(x)} \cdot g'(x).$$

In your final example you have functions $f:\mathbb R^5\to\mathbb R$, $u,v:\mathbb R^2\to\mathbb R$, $g:\mathbb R\to\mathbb R$. Now $f(x,y,g(x),u(x,g(x)),v(x,g(x))$ is a real number. If you want to "variate the $x$ that appears as argument of $g$", you should write $$\left.\frac{\partial}{\partial a}f(x,y,g(a),u(x,g(a)),v(x,g(a))\right|_{a=x},$$ If you want to "variate the $x$ that appears as first argument of $u$", then write $$\left.\frac{\partial}{\partial a}f(x,y,g(x),u(a,g(x)),v(x,g(x))\right|_{a=x}$$ and if you want to "variate the $x$ that appears as first argument of $v$", then write $$\left.\frac{\partial}{\partial a}f(x,y,g(x),u(x,g(x)),v(a,g(x))\right|_{a=x}.$$

Anyway

Leibniz notation is very susceptible to misinterpretations and ambiguities just like the ones you have pointed out. I think one big reason for this is that it conflicts with functional notation: variables in parentheses denote dependence instead of evaluation.

Another problem is that unfortunately there seem to be different ways to use this notation and I haven't found a single place where they explicitly tell the relation between their convention and functional notation. E.g., the interpretation I gave above corresponds to $$\frac{\partial}{\partial x_i}f(x_1,\dots,x_n) = (t\mapsto f(x_1,\dots,x_{i-1},t,x_{i+1},\dots,x_n))'(x_i).$$ For these reasons I usually tend to discourage the use of Leibniz notation. If I want to denote the derivative of the function $f$ with respect to its $i$th entry, I simply write $D_if$ and I think it's perfectly clear.

Hope this helps.

Jackozee Hakkiuz
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  • In your composition notation then, we lose the information that $x$ and $a$ are the same algebraic variable, as a tradeoff for highlighting the fact that the calculus is viewing them differently, correct? As for the non-Leibniz notation, it does remove one layer of ambiguity, but can it fully disambiguate the examples in the last paragraph of my question (if these are valid partial derivatives with respect to $x$ in the first place, which is the main thrust of my question)? – user10478 Apr 23 '19 at 18:09
  • @user10478 what exactly do you mean by "the same algebraic variable"? If one is varying and the other isn't, then they are not the same. Or are you referring to the fact that you want them both to represent the same quantity-measured-in-the-real-world? that has nothing to do with algebra, and all to do with interpretation.

    And yes, the operator notation can fully disambiguate every case of partial derivative you need to use - as well Lebniz notation, when used properly.

     – Jackozee Hakkiuz Apr 25 '19 at 16:21
  • As for the "partial derivatives with respect to x" business: don't take me wrong, but I think there is a little misconception. One doesn't take "partial derivatives with respect to a variable". One takes "partial derivatives with respect to one entry of a function". The name you give to the variable that particular entry eats is completely accidental. – Jackozee Hakkiuz Apr 25 '19 at 16:28
  • I was thinking that "varying" really has to do with adding $h$ in the formal limit definition, i.e., in $\frac{f(x + h,\ y,\ z(x)) - f(x,\ y,\ z(x))}{h}$, every occurrence of $x$ is the same $x$. It's just that in applying this particular meaning of the partial derivative, the first parameter of $f$ gets run through the "add h" function, while the parameter of $z$ does not. Replacing one of the $x$'s with $a$ seems to discard this information. – user10478 Apr 25 '19 at 17:13
  • @user10478 Hey. Very likely you already solved it yourself, but I kept thinking about your question. I modified my answer. – Jackozee Hakkiuz Aug 02 '20 at 01:45
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Here we look at a convenient setting for the current situation which shows there is no disambiguity when doing the partial derivative.

We consider a differentiable multivariate function \begin{align*} &f:\mathbb{R}^3\to\mathbb{R}\\ &(u,w,z)\to f(u,w,z) \end{align*} and we let the variables $u,w,z$ be real-valued differentiable functions in $x$ \begin{align*} &x\to u(x)\\ &x\to w(x)\\ &x\to z(x) \end{align*}

This is a general setting for the current case \begin{align*} f(x,y,z(x)) \end{align*} where

  • $u(x)=x$ is the identity function,

  • $w(x)=y$ is the constant function $y$ and

  • $z(x)$ is a real-valued differentiable function in $x$.

In order to get $\frac{\partial}{\partial x} f(x,y,z(x))$ we calculate the total derivative \begin{align*} \frac{\partial}{\partial x}&f(u(x),w(x),z(x))\\ &=f_uu^{\prime}(x)+f_ww^{\prime}(x)+f_zz^{\prime}(x)\tag{1} \end{align*}

Let's look at an example, for instance \begin{align*} f(u,w,z)&=3uw+z^2\\ u(x)&=x\\ w(x)&=y\\ z(x) \end{align*}

We calculate the total derivative according to (1) and obtain \begin{align*} \frac{\partial}{\partial x}&f(u(x),w(x),z(x))\\ &=f_uu^{\prime}(x)+f_ww^{\prime}(x)+f_zz^{\prime}(x)\\ &=3w(x)u^{\prime}(x)+3u(x)w^{\prime}(x)+2z(x)z^{\prime}(x)\\ &=3y\cdot 1+3x\cdot 0+2z(z)z^{\prime}(x)\\ &\,\,\color{blue}{=3y+2z(x)z^{\prime}(x)}\tag{2} \end{align*} On the other hand we replace all arguments $u,w,z$ of $f$ by the underlying functions in $x$ and obtain \begin{align*} \frac{\partial}{\partial x}&f(u(x),w(x),z(x))\\ &=\frac{\partial}{\partial x}\left(3u(x)w(x)+\left(z(x)\right)^2\right)\\ &=\frac{\partial}{\partial x}\left(3xy+\left(z(x)\right)^2\right)\\ &\,\,\color{blue}{=3y+2z(x)z^{\prime}(x)}\tag{3} \end{align*}

We observe (2) and (3) coincide showing how to calculate the partial derivative.

When looking at a more complex function like \begin{align*} \frac{\partial}{\partial x}f(x,y,z(x),a(x,z(x)),b(x,z(x))) \end{align*} we can do the calculation in the same way as above, namely considering

\begin{align*} &f:\mathbb{R}^5\to\mathbb{R}\\ &(u_1,u_2,u_3,u_4,u_5)\to f(u_1,u_2,u_3,u_4,u_5)\\ &u_1(x)=x\\ &u_2(x)=y\\ &u_3(x)=z(x)\\ &u_4(x)=a(v_1(x),v_2(x))\\ &\qquad v_1(x)=x\\ &\qquad v_2(x)=z(x)\\ &u_5(x)=b(w_1(x),w_2(x))\\ &\qquad w_1(x)=x\\ &\qquad w_2(x)=z(x) \end{align*} and continuing similarly as above, although the calculations are now admittedly somewhat more cumbersome.

Markus Scheuer
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If you paremetrize z in terms of x, $z(x)$, then $f(x,y,z(x))$ is actually a function depending on x & y, $f(x,y)$. taking a $\partial _xf(x,y)=\partial _xf(x,y,z(x))$ if you define $\partial _xf(x,y,z(x))$ as "taking derivative with respect to (wrt) x and not z(x)" + $\partial _zf(x,y,z(x))z'(x)$. But after all, I think it is a convention, since defining $\partial_x$ as "taking derivative wrt x and not z(x)" can be useful.

Unknown
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