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I am going through Tenenbaum and Pollard's book on differential equations and they define the differential $dy$ of a function $y = f(x)$ to be the function $$ (dy)(x,\Delta x) = f'(x) \cdot (d\hat{x})(x, \Delta x) $$ where

  • $\Delta x$ is a variable denoting an increment along the $x$-coordinate
  • $\hat{x}$ denotes the function $\hat{x}(x) = x$, and
  • $d\hat{x}$ is the differential of the function $\hat{x}$.

I've never seen differentials crisply defined this way. They're usually described as "small quantities" or just avoided in favor of definitions of the derivative in terms of limits. Anyway, this definition makes good sense to me. Is this the accepted way to think of them -- i.e. as functions?

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Yes, differentials and derivatives are 2 ways of doing the same thing to a function.

The derivative operator $\frac{d}{dx}$ hits functions and $\frac{d}{dx}f$ is itself a function. The differential operator $d$ hits functions and $df$ is itself a function. Therefore with this definition of $df$, you can think of derivatives as fractions because the fraction $\frac{df}{d\hat{x}}$ is precisely the derivative $\frac{d}{dx}f$

The only difference is that $\frac{d}{dx}f$ has the interpretation of a rate of change, while $df$ has the interpretation of the change in height of the tangent line.

Note that the function $\hat{x}$ is also called the identity function. He goes on to say that over time $d\hat{x}$ turned into $dx$. And that we have to remember that $dx$ stands for the differential of the identity function (remember it's only functions that we can take differentials of). However the identity function is the one function in which the word differential becomes synonymous with increment/$\Delta x$ (large or small)/infinitesimal.

Notice that when you are asked to find the antiderivative of a function, the symbol $\int \dots dx$ is used. Therefore $\int g(x) \;dx$ is asking for a function $G$ whose $G' = g$. However, imagine grouping $g(x)dx$ together, leaving $\int$ by itself. $g(x)dx$ takes the form of a differential of some other function. So, in order to get the same answer as with antiderivatives, define $\int$ as the operator that produces the anti-differential of whatever follows $\int$. That function is still $G$! Because $dG = g\;dx$

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  • $\begingroup$ In other words, when you have differentials of independent variables, the differential secretly stands for the differential of the corresponding identity function (it's only functions which you can take differentials of). It's only for these functions which differential and infinitesimal/increment in the independent variable can be used interchangeably. If you randomly see a $dy$. Then either there is a function of $y$ and $y$ is the independent variable. Or equivalently, there is an implicit function between $y$ and $x$.Then $dy$ is equivalently the differential of the function relationship $\endgroup$ – DWade64 Oct 18 '18 at 13:57
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    $\begingroup$ Thanks. Actually, there is something ambiguous happening in the notation here. Let $y = f(x)$. In this context, $dy$, means the function $(dy)(x, \Delta x) = f'(x)\Delta x$. But if $y = f(x(t))$, then $dy$ refers to the different function: $(dy)(t, \Delta t) = f'(x(t)) \cdot (dx)(t, \Delta t)$. So $dy$ is notationally vague. I suppose we infer the right meaning by context. Regarding Tennenbaum's claim that $d\hat{x}$ gradually became $dx$ over time, I doubt this account is accurate. More likely, $dx$ was the original notation and folks figured out it meant $d\hat{x}$ later on. $\endgroup$ – ted Oct 18 '18 at 22:17
  • $\begingroup$ @ted Yeah I never took that point seriously. Even further, $y$ is not the function. $f$ is the function. Therefore $dy$ is always ambiguous. Consider an xy plane with the graphs $y = f(x)$ and $y = g(x)$. The 2 functions are $f$ and $g$. But $dy/dx$ is ambiguous now. In my answer, I tried to avoid talking about dy altogether. Im actually starting to like differentials more and more over derivatives. Yeah my comment about two different functions $f$ and $g$ is exactly the same as your comment about $f(x)$ and $f(x(t))$. These are 2 completely different functions. One is $f$ and the other $\endgroup$ – DWade64 Oct 18 '18 at 22:22
  • $\begingroup$ is the composition $f\circ x$. Composition functions should always be given another name because they are totally different functions (or constantly write $f(x(t))$. Because of this (not giving another name to the composition), textbooks everywhere often write the chain rule wrong. But you make a very good point $\endgroup$ – DWade64 Oct 18 '18 at 22:29
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Yes, the differential is a linear map in terms of its second argument, i.e. it is proportional to the increment. The constant of proportionality depends on $x$ and is the local value of the derivative of $f$.

For these reasons, you have

$$d\hat x(x,\Delta x)=\Delta x.$$

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