# Is this example of Leibniz notation sloppy?

In helping a family member who is studying calculus, I was asked about the meaning of the following, which is straight out of a calculus text book (Varberg and Purcell)

$v'(t) = \frac{{\rm d}\,v}{{\rm d}t}$

I've also seen the same thing elsewhere.

Now I haven't done much calculus since my undergrad, but I've done a lot of logic and C.S.; so much, in fact, that I am now unable to read a formula without fretting about the type and binding of each variable. To my (perhaps hypercritical) eye this looks plain wrong: on the, left $v$ is being treated as a function. On the right, $v$ is being treated as a (dependent) real variable. To me, it seems the equation should be written as

$v'(t) = \frac{{\rm d}\,v(t)}{{\rm d}t}$

or as

$\dot{v} = \frac{{\rm d}\,v}{{\rm d}t}$ ,

depending on whether you want to treat $v$ as a function or as a variable.

My question: Is the text book being sloppy about notation, or is there some defensible interpretation of $v'(t) = \frac{{\rm d}\,v}{{\rm d}t}$ that I'm missing?

• There is some related discussion at matheducators.stackexchange.com/questions/562/… . But I don't think my specific question is addressed. Aug 26 '14 at 14:15
• I agree with you. $v'(t)$ is a number. $\frac{dv}{dt}$ is a function. It doesn't make sense to say they're equal. People are often sloppy about this though, hoping that the correct interpretation is clear from context. I also agree with your proposed alternatives. Sep 2 '14 at 10:22
• @littleO: how could $\frac{dv}{dt}$ possibly be a function from $\mathbb{R}\to\mathbb{R}$? If so then $v$ must also have been a function, but then it makes no sense to mention the dependent variable in the notation of the derivative. It's amazing how much confusion exists about this notation. Oct 24 '16 at 8:06
• @MichaelBächtold Yes, $v$ is a function, and $\frac{dv}{dt}$ is (by definition) just another way of writing the function $v'$. I agree that it's a little weird for a letter $t$ to appear in the name of this function. (Of course, this notation is supposed to remind us that $v'(t) \approx \frac{\Delta v}{\Delta t}$, where $\Delta t$ is a tiny change in the value of $t$ and $\Delta v$ is the corresponding change in the value of $v$ (as its input changes from $t$ to $t+\Delta t$).) If there's any confusion about Leibniz notation, it's always possible to avoid Leibniz notation entirely. Oct 24 '16 at 8:52
• @littleO Well, I'm not in favor of abandoning Leibniz Notation, since it has to be used when applied to a variable. While the prime should be used only applied to a function. Maybe this answer explains better what I'm trying to say math.stackexchange.com/a/1259113/1984 Oct 24 '16 at 8:56

Tl;dr: They are all equal and this is a matter of preference. The notations mean the same concept, each with its respective advantages and disadvantages.

Newton's notation $\dot{v}$ is commonly used in physics and will always mean you are considering a function of time, denoted by the parameter $t$. This notation has the advantage that you avoid confusion regarding with respect to what is the function being differentiated. The disadvantage is that this is one of the cases of frozen notation: don't dare writing $\dot{v}$ for differentiation with respect to anything else but $t$.

Leibniz's notation $\frac{dv}{dt}$ is very flexible. The advantages of it are that it already tells you what is the function differentiated and with respect to what. It also adapts easily to all cases, as this is not frozen notation. The disadvantage is that there is the slight abuse you have noticed. Correctly we should write $\frac{dv(t)}{dt}$, but $\frac{dv}{dt}$ conveys the idea pretty well.

Lagrange's notation $v'$ is the most confusing of all, in my opinion. Is not entirely clear what you are differentiating with respect to. Sure, if you have a one variable function it works, but I've seen people write $v'$ when $v$ is a function of space and it meant differentiation with respect to time, without mentioning it. If abused it can lead to great trouble. The chain rule in this notation states that $(f(g(x))' = f'(g(x)) g'(x)$, but what if you omit the parameters? It becomes $(f \circ g)' = f' g'$, which is horrible (it may even confuse students with the product rule).

• So you agree it's an abuse of notation, although "slight". Also: shouldn't the point-free chain rule be $(f \circ g)' = (f' \circ g)\,g'$ ? Aug 27 '14 at 13:17
• @TheodoreNorvell Yes, but if you omit the function $g$ inside $f'$ you get what I write. Physicists always do it: just see how they apply the chain rule for multivariable functions. Aug 27 '14 at 13:53
• @MarkFantini A physicist would write $df/dx = df/dg \., dg/dx$, which while it does lead to issues in multiple variables is not nearly as broken as what you wrote. You are actually using the prime in two different ways in the same expression, while they simply leave the composition implicit but possible to guess from context. Aug 27 '14 at 19:51
• Mark Fantini, further to @RyanReich's remark, if one suppresses the 'inside' function, then isn't one forced to do so both before and after differentiating, and so write the even more horrible $f' = f'g'$? Sep 1 '14 at 1:25

I disagree with your descriptions of the meaning of the various uses of $v$; in fact, I would flip them! I think that what I say below describes the common mathematical perspective, but it could be that I am wrong; for example, Mark Fantini above (https://matheducators.stackexchange.com/a/4282/2070) seems to disagree.

In mathematical convention (which differs from some computer-science conventions), a 'bare' name $v$ refers to a function, whereas an 'argumented' name $v(t)$ refers to its value. It is very, very common to abuse this, and so to refer to "the function $\sin(x)$" rather than "the function sin" (which looks terrible, or perhaps theological) or "the function $x \mapsto \sin(x)$", especially in contexts where the function has no name (like "the function $x^2$").

Thus, I would argue that combining convention with precision (which is not necessarily desireable!) would dictate writing $v' = \frac{\text dv}{\text dt}$ (or $\dot v$ if you prefer it in place of $v'$), or, if you really want, the cumbersome and fairly ugly notation $v'(t) = \frac{\text dv}{\text dt}(t)$. (Another common approach to indicating a value to be plugged into a derivative is to write something like $\left.\frac{\text dv}{\text dt}\right|_{t = 7}$ for $v'(7)$.)

I find an approach that rarely mentions explicit arguments particularly valuable in a differential-equations course, where, for example, we might have a height $y$ depending on a position $x$, which in turn depends on a time $t$—in which context it is much easier to get to the bottom of the physical meaning if one writes just $\frac{\text dy}{\text dx}$ or $\frac{\text dy}{\text dt}$ rather than, say, $\frac{\text dy(x)}{\text dx}$ or $\frac{\text dy(x(t))}{\text dt}$.

• The way I see it, $\frac{{\rm d}}{{\rm d}t}$ is an operator that applies to real expressions rather than functions. That is why I said that in $v'(t) = \frac{{\rm d}\,v}{{\rm d}t}$ it looks like $v$ is being treated as a function on the left and a variable on the right. Your suggestion of $v' = \frac{{\rm d}\,v}{{\rm d}t}$ suggests that $\frac{{\rm d}}{{\rm d}t}$ is properly applied to functions and that, say, $\frac{{\rm d}\,t^2}{{\rm d}t}$ is just sloppy notation for $\frac{{\rm d}}{{\rm d}t}(\lambda t.t^2)$. This is an interesting idea, but it seems at odds with common practice. Sep 1 '14 at 1:01
• @TheodoreNorvell, what you've written is a perfect summation of what I meant to say, which, I believe, is what most mathematicians would think if asked formally to explain the derivative notation—even though it clashes with practice. (See also the difference between how a mathematician defines a proof and how a mathematician writes a proof!) For example, I think that every mathematician would say (even if it conflicts with practice) that the domain of $\text d/\text dt$ is a space of functions, not of formulæ. Sep 1 '14 at 1:23
• @L Spice, thanks. Perhaps I should have said "$\frac{{\rm d}\,t^2}{{\rm d}t}$ is just sloppy notation for$\frac{{\rm d}\,(\lambda t,t^2)}{{\rm d}t}(t)$." Sep 2 '14 at 11:25