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I'm trying to understand Leibniz notation. I have read the Wiki page on the notation and numerous others, but think I'm overlooking something crucial to understanding.

I have been told that:

$$ y=f(x) $$ $$ f'(x) = \frac {dy}{dx} $$

Question 1: In the above form, I'm assuming the strings of symbols $x$ (LHS) and $dx$ (RHS) should be treated as different objects, and that $dx$ represents both the denominator of the difference quotient $\lim_{h\rightarrow0}{\frac {f(a+h) - f(a)}{(a+h) - a}}$ at some point $a$ and that $x$ is the independent variable which to infinitesimally change. Is that correct? To eliminate symbol confusion, is the following form equivalent?

a.

$$ f'(a) = \frac {dy}{dx} \Big|_{a}= \frac {dy}{dx} = \lim_{h\rightarrow0} {\frac {f(a+h) - f(a)} {(a+h) -a}} $$ An example of the chain rule:

b. $$ f(x) = (f_1 \circ f_2 \circ f_3)(x) $$ $$ y = f_1(u), u = f_2(v), v = f_3(x) $$ $$ f'(a) = \frac {dy} {du} \Big|_{f_2(f_3(a))}\frac {du} {dv} \Big|_{f_3(a)} \frac {dv} {dx} \Big|_{a} $$

$$ \begin{align*} f'(a) &= (f_1' \circ (f_2 \circ f_3))(a) \cdot (f_2' \circ f_3)(a) \cdot f_3'(a) \end{align*} $$

Question 2: I understand the vertical bars $\Big|_{z}$ provide a point from which to take the derivative. I observe there is notation with and without bars and have not seen any history of Leibniz using them. I'm assuming they help reduce back references to the function definitions. Does anyone know why used?

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  • $\begingroup$ These notations are equivalent: $$f'(a) \equiv f \mid_a$$ $\endgroup$
    – vitamin d
    Commented Feb 26, 2021 at 3:31
  • $\begingroup$ What about $\frac {df} {dx}$? I don't see how it can be used $\endgroup$
    – Nick
    Commented Feb 26, 2021 at 3:50
  • $\begingroup$ Oh sorry. I made a mistake in my comment. The right hand side should of course be $$\frac{d}{dx}f\mid_a.$$ $\endgroup$
    – vitamin d
    Commented Feb 26, 2021 at 3:51
  • $\begingroup$ @vitamind Ok, that seems to follow the post. I'm confused why I see material with and without the bar notation. Also, it seems like it's required to really proceed with the chain rule using Leibniz notation. $\endgroup$
    – Nick
    Commented Feb 26, 2021 at 3:58
  • $\begingroup$ $f'(a)$ is the notation most people go with. (if $a$ is a constant). $\endgroup$
    – vitamin d
    Commented Feb 26, 2021 at 4:00

1 Answer 1

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I think to understand Leibniz notation it is best to get into a “Leibniz mindset” because at his time the idea of a function was different from today. As far as I know, he would talk about $y$ being a a function of $x$, meaning that the value $x$ determines the value of $y$. In modern parlance: There exists a (modern) function $f : \mathbb{R} \to \mathbb{R}$ such that $y = f(x)$.

But note that there is no $x$ on the left-hand side. The value of $y$ really depends on the “current” value of $x$. This is different from modern functions where writing $f(x) = x^2$ and $f(z) = z^2$ really defines the same function $f$. As such, writing $y(x_0)$ does not really make sense, as $y$ might be a function of different variables in different ways, and to express the same thing one should specify which variable one fixes and write $y|_{x=x_0}$ (i.e. mention $x$ explicitly). Writing $y|_{x_0}$ or even $y(x_0)$ is a common shorthand, though, and often does not lead to confusion.

In this context, Leibniz notation makes sense: An (infinitesemal) change of $x$ gives by the implicit dependence of $y$ on $x$ an (infinitesemal) change of $y$; their quotient is the derivative $\frac{dy}{dx}$. Now as you know, the value of this also depends on the value of $x$. In other words, $\frac{dy}{dx}$ is again a “function of $x$”, namely $\frac{dy}{dx} = f'(x)$ (if $y = f(x)$).

If we want the value of this at a specific point, we can write $\frac{dy}{dx}|_{x = x_0}$ or $\frac{dy}{dx}|_{x_0}$ for $f'(x_0)$.

For Leibniz' chain rule, we are in the following situation: The variable $u$ is a function of $x$, say $u = f(x)$, and the variable $y$ is a function of $u$, say $y=g(u)$. Substituting, we see that $y = g(f(x)) = (g \circ f)(x)$, i.e. $y$ is also a function of $x$. We can therefore try to compute $\frac{dy}{dx}$. Using the chain rule you know for the $'$-notation, we see: $$ \frac{dy}{dx} = (g \circ f)'(x) = g'(f(x)) f'(x) = g'(u) f'(x) = \frac{dy}{du} \frac{du}{dx}. $$ This is the usual form of this chain rule. Note that the first factor $\frac{dy}{du}$ is a function of $u$ but also a function of $x$ because $u$ is a function of $x$.

If we want to evaluate at a specific point $x_0$, we run into the problem mentioned above that $\frac{dy}{du}$ is a function of both $x$ and $u$ in different ways. Now $x_0$ looks like is should be the value for $x$, but in your question you use $a$ where this is less clear. And $\frac{dy}{du}$ is more obviously dependent on $u$; so to be explicit, we write down the transformation from $x$ to $u$, i.e. $$ \frac{dy}{dx}|_{x=x_0} = \frac{dy}{du}|_{u=f(x_0)} \frac{du}{dx}|_{x=x_0}. $$ Using the shorthands above, $$ \frac{dy}{dx}|_{x_0} = \frac{dy}{du}|_{u(x_0)} \frac{du}{dx}|_{x_0}, $$ where we also employed the confusion between $u$ and the function $f$.

Why do we use this notation? Aside from convention, it has two advantages:

  1. Defining a dependence is often easier, i.e. we can write down and derive the term $x^3 + \exp(x)$ immediately instead of having to define a function $f$ with $f(x) = x^3 + \exp(x)$ first.

  2. Once you have functions of several variables it is nice to be able to refer to them by name in the derivative notation (instead of having to use numerical indices). I think this is why physicists often still use this notion of “function”.

Last, I want to mention that if you go on to study differential geometry, you will learn how to do all of this in a way that kind of gives you the best of both worlds: Normal function (which I think are conceptually much clearer than these “functions of $x$”) combined with the ability to refer to names instead of indices.

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  • $\begingroup$ Is this the correct understanding? Confused over the argument and non-argument form: $y = f(x)$,$\frac {dy}{dx} = f'$,$\frac {dy}{dx} \Big|_{x} = f'(x) = \lim_{h \to 0} {\frac {f(h+x) - f(x)}{(h+x) -x}} $ $\endgroup$
    – Nick
    Commented Feb 27, 2021 at 15:37
  • $\begingroup$ I’m pretty sure that my explanation of Leibniz' understanding of the term “function” is correct and at least for me, this helps interpreting Leibniz notation. I would not write $\frac{dy}{dx} = f'$; the left-hand side is a term (that depends on $x$, i.e. a “function of $x$”) whereas the right-hand side is a (modern) function (no “of $x$” here). To make this correct, you have to supply $x$ on the right, i.e. $\frac{dy}{dx} = f'(x)$. $\endgroup$ Commented Feb 27, 2021 at 16:50
  • $\begingroup$ I also think that $\frac{dy}{dx}|_x$ is strange. I interpret this as “substitute $x$ for $x$ in the term $\frac{dy}{dx}$” which of course does not change anything. $\endgroup$ Commented Feb 27, 2021 at 16:53
  • $\begingroup$ It is kind of saying: “I’m fine with $y$ having this implicit dependence on $x$ (so I can compute $\frac{dy}{dx}$) but once I have done that, implicitly depending on $x$ is evil so I better make the dependence explicit by adding $|_x$).” $\endgroup$ Commented Feb 27, 2021 at 17:02
  • $\begingroup$ My observations so far is the bar notation reduces back referencing the function definitions, which are the same form if taken on the bottom of the bar. $\endgroup$
    – Nick
    Commented Feb 27, 2021 at 17:05

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