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I have on a textbook a system of two ODEs as follow

$$\dfrac{df}{dx}=af(x)g(x)\\ \dfrac{dg}{dx}=-bg(x)$$

So, the author studies $df/dg$ writing $\dfrac{df}{dg}=-\dfrac{a}{b}f(x)$.

I understand the arithmetic, but I am not sure of mathemathical/geometric meaning.

Thank you in advance for light and maybe some lemmas or theorems of reference that it is valuable (are when is valuable).

Quiet_waters
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2 Answers2

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I have ranted numerous times on this site about that horrendous notation. It really means nothing and can be very misleading. I would suggest you write $$\frac{df}{dx} = af(x)g(x) = -\frac ab f(x)\frac{dg}{dx},$$ and then divide by $f(x)$ to get $$\frac1{f(x)}\frac{df}{dx} = -\frac ab \frac{dg}{dx}.$$ Now integrate both sides (with respect to $x$) and you get $$\ln |f(x)| = -\frac ab g(x) + c.$$ [You could write the previous equation as $\frac{df}f = -\frac ab dg$, and then you get where your textbook should end up.]

Ted Shifrin
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    For better or for worse, this is a very common notational abuse. For example, the chain rule is often written as $$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} , ,$$ where you are differentiating $y$ with respect to the function (!?) $u$. – Joe Dec 30 '20 at 20:33
  • @TedShifrin Yes, although I have changed my profile, I was one of the person that you bicker about this confused notation (rs). However, this time, I take the notation in a very consecrated article (two very consecrated articles in fact, that I just rewrote to post as question). As Joe said, this is a very common notational abuse, but I'd like to thank you so much for show an alternative way that avoid it! Thank once again, Prof. Shifrin. – Quiet_waters Dec 30 '20 at 20:37
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    @Joe: No, when we write the chain rule, we really have $y=f(u)$ where then we substitute $u=g(x)$, so $u$ is the independent variable for $f$, not a function. I did teach calculus for mire than 45 years, and it can be done in a way that makes sense. – Ted Shifrin Dec 30 '20 at 20:52
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    @TedShifrin That's what I meant—the usual formula is an abuse of notation because it makes it seem as if you are differentiating a function with respect to a function. If $y=f(g(x))$, and $u=g(x)$, then $$\frac{dy}{du} = \frac{df(g(x))}{dg(x)} , .$$ However, the RHS is 'overloading' the function $g(x)$. It is being used both as a dummy variable to illustrate that we are talking about $f'$, and also as the point at which $f'$ is being evaluated. Hence, we could write this is more clearly as $$\frac{df(u)}{du}\Biggr|_{u=g(x)} , .$$ – Joe Dec 30 '20 at 20:57
  • I discovered the question was being discussed here https://math.stackexchange.com/questions/291376/differentiate-with-respect-to-a-function and here https://math.stackexchange.com/questions/954073/derivative-of-a-function-with-respect-to-another-function – Quiet_waters Dec 30 '20 at 21:09
  • @TedShifrin - Hi Dr. Shifren, I have a quick question pertaining to the portion of your answer regarding notation: what are your thoughts on writing $\frac{df}{dx}(x)$ for the derivative of a function $f$ evaluated at a point $x$ (example using this notation: given $f(x)=x^{2}$, then $\frac{df}{dx}(x) = 2x$)? – Taylor Rendon Oct 18 '22 at 01:40
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    @MathDoctor That's fine, although basically no one does it. I personally prefer the non-Leibniz notation $f'(x)$ because it makes it explicit where you're evaluating the function $f'$. The sorts of things we write like $$\frac{dx}{dy} = \frac1{\frac{dy}{dx}}$$ are horrendous, because we completely suppress where the derivatives are evaluated (so kudos to you for wanting to make it explicit). – Ted Shifrin Oct 18 '22 at 01:44
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This notation is often difficult to make sense of. What, for instance, would it mean to differentiate $\sin(x)$ with respect to $\cos(x)$? To begin with, we must write $\sin(x)$ in terms of $\cos(x)$: $$ \sin(x)=\sqrt{1-\cos(x)^2} \, , $$ ignoring the $\pm$ signs that should be on the square root for now. This means that $$ \frac{d\sin(x)}{d\cos(x)}=\frac{d\sqrt{1-\cos(x)^2}}{d\cos(x)} \, . $$ But this is the same as $$ \frac{d}{du}\left(\sqrt{1-u^2}\right)\Biggr|_{u=\cos(x)} \, . $$ So every time you see $$ \frac{df(x)}{dg(x)} \, , $$ note that $f(x)=h(g(x))$ for some function $h$, and write $$ \frac{dh(g(x))}{dg(x)}=\frac{dh(u)}{du}\Biggr|_{u=g(x)} \, . $$ The $u$ is a dummy variable that can then be replaced by whatever function you are 'differentiating with respect to'. You can use $g(x)$ as a dummy variable, but it often leads to confusion, and is considered by many people to be poor style.

To give another example, let's consider the chain rule, $$ \frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx} $$ Say $y=\sin(u)$, where $u=x^2$. Then, the chain rule appears to say $$ \frac{d}{dx}(\sin(x^2))=\frac{d}{dx^2}\left(\sin(x^2)\right) \cdot \frac{d}{dx}(2x) \, . $$ It seems difficult to parse the first term on the RHS. The easiest way to do this is to write it instead as $$ \frac{d}{du}(\sin(u)) \Biggr|_{u=x^2} \, . $$ However, we can also treat $x^2$ just as we would any other variable, and make sense of the term that way, too.

Joe
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  • Joe, I accepted Prof. Shifrin's answer and would like to bount your answer as well as soon as system liberate. But I'd like to confess you a confusion that hurt my mind. And if the function $g(x)$ is not one-to-one, this "variable" could lead to some mathematical confusion maybe?! Thank you. @TedShifrin – Quiet_waters Dec 30 '20 at 20:41
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    @Quiet_waters I wouldn't worry too much about it—I think your comment has demonstrated how imprecise Leibnizian notation can be. If I can find a way to make sense of $df/dg$, then I'll edit my post. – Joe Dec 30 '20 at 21:00
  • Yes Thank you @Joe – Quiet_waters Dec 30 '20 at 21:06
  • I discovered the question was being discussed here https://math.stackexchange.com/questions/291376/differentiate-with-respect-to-a-function and here https://math.stackexchange.com/questions/954073/derivative-of-a-function-with-respect-to-another-function – Quiet_waters Dec 30 '20 at 21:09
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    @Quiet_waters: My feelings won't be hurt if you accept Joe's answer instead. There are really two issues going on here. One is the Leibniz notation and its abuse; the other (which was more what I cared about here) is the usual separation of variables game for differential equations (which has been discussed numerous times on here, too), where I have argued before that it's really just an application of the chain rule to justify it. – Ted Shifrin Dec 30 '20 at 21:31
  • Not at all @TedShifrin. I did not make the comment in this regard. My only intention was to accept both answers (but it is impossible). So I will accept you (because I think it got to the heart of my dissatisfaction with the notation) but also evaluate Joe's (because it in some way gives sense to the notation). The question in fact was about the care of the notation abuse as you pointed out. Thank you all – Quiet_waters Dec 30 '20 at 21:53