Is this the correct interpretation of the differential?

Question

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?

Answer 1

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 $dG = g\;dx$

Answer 2

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

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