Partial Derivative Disambiguation

Question

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 $f$?

Answer 1

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

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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 $$\left.\frac{\partial}{\partial a}f(x,y,g(x),u(x,g(x)),v(a,g(x))\right|_{a=x}.$$

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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.

Answer 2

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.

Answer 3

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.