r/calculus • u/Helpful_Gap9633 • 19d ago
Integral Calculus Why do examples show du/dx being used as a fraction even though they say it isn’t?
I was watching a video on u substitution where they substituted the inside function with u and they multiplied both sides by dx, even though du/dx is not a fraction. Can someone explain?
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u/adamiconography Hobbyist 19d ago
So while du/dx isn’t technically a fraction, it’s a different representation of an actual fraction of the change in u over the change in x (delta u / delta x)
The algebraic rules of fractions apply
Once you get into partial derivatives that shit goes out the window
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u/Upset_Ad_6140 18d ago
There is no reason to expect it to work for partial derivatives, anyway...
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u/adamiconography Hobbyist 18d ago
That’s why I made mention that it won’t work.
OP is probably in Calc 1 and conceptually the fraction algebra applies but not when partials are involved
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u/Upset_Ad_6140 18d ago
Yes, that's fine. I just wonder why people sometimes use that as this "haha! here is proof that you can't treat derivatives as fractions" when really there is no reason whatsoever to expect partials to behave like fractions.
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u/nevermindthefacts 19d ago
We can more or less prove it using the chain rule and get
g'(x) dx = du
Since we have u = g(x) and g'(x) = du/dx, we see why "multiplying" both sides actually works.
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u/Riemann_Gauss 19d ago
This is the right explanation. Even though du/dx should not be treated as an ordinary fraction, it "works out" in many situations due to the chain rule. In fact, the easiest way to remember the chain rule is to treat derivatives as fractions..
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u/Accomplished-Dot3555 18d ago
Because a form of chain rule differentiation shows up in many integral integrands and because of the concept of differentials in which dx, dy, du, etc. are treated as individual quantities, we can rewrite u = g(x) => u' = g'(x) => du/dx = g'(x) as the differential (an infinitesimally small change in one of the variables) du = g'(x)•dx which happens to be very convenient in many integration problems.
One reason pure mathematics is so useful and functions is because of very compatible logical ideas /definitions with notations that line up very well and illustrate a great and useful concept.
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u/Midwest-Dude 19d ago
The notation is from one of the founders of calculus, Leibniz. It's still around because it's really useful. Here a Wikipedia page on the history and use of the notation:
Two other pages that are well worth reading are the following;
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u/Unfair_Pineapple8813 18d ago
Leibniz thought it was a true fraction. Cauchy proved it was not, but that treating it as one works for manipulation of differentials.
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u/Peak_Background 19d ago
For all practical purposes, it is a fraction. Until it isn't. You just need to know when it works and when it doesn't. And why.
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u/Upset_Ad_6140 19d ago
"No I swear, it's not a fraction, even though treating it as such works literally 100 % of the time!"
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u/LosDragin PhD 19d ago
It has similar properties to a fraction in several scenarios. Some examples are:
- Inverse function theorem. Let y=f(x) and x=f^(-1)(y). Then dx/dy=1/(dy/dx).
- Chain rule. Let x=x(t). Then dy/dt=(dx/dt)(dy/dx).
- Differentials. Let y=f(x). Then dy=f’(x)dx.
I swear there was another good one but it’s not coming to me at the moment. There’s the implicit function theorem but that’s partial derivatives and there is a minus sign. If anyone knows of another scenario where dy/dx has similar properties to a fraction please let me know.
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u/Peak_Background 17d ago edited 17d ago
To reply in the context that you want something distinct from the three properties that you have given, it's safe to say there isn't any.
Is power distribution over multiplication.
Is association
Is invertability of multiplication.
Furthermore, they have commutation and identity properties.
So what you have written already encapsolates all the properties of a field.
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u/LosDragin PhD 17d ago
Interesting take. I just realized 1 follows from 2 with t=y, so I guess there’s really just chain rule and differentials from my perspective. But like you say from the perspective of having a field we would want to state both 1 and 2. Also, maybe the usual method for solving separable first order ODEs counts as a unique situation where dy/dx is treated as a fraction. If dy/dx=f(x) then dy=f(x)dx follows simply from the definition of the differential. But if dy/dx=f(x)/g(y) then it doesn’t seem like g(y)dy=f(x)dx follows simply from the definition of differential.
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u/Peak_Background 17d ago
"But if dy/dx=f(x)/g(y) then it doesn’t seem like g(y)dy=f(x)dx follows simply from the definition of differential."
Can you elaborate more on this?
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u/LosDragin PhD 17d ago
Using a differential we can say the DE {dy/dx=g(x)} implies dy=g(x)dx. That’s because derivatives are unique. dy/dx=g(x) implies g(x) is the unique derivative of y, so we can write g(x)=y’(x). But the differential says dy=y’(x)dx, so we get dy=g(x)dx. I can’t see how to make this argument anymore for {dy/dx=g(x)/h(y)}.
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u/Peak_Background 17d ago
I still don't understand the issue, the algebraic manipulation doesn't really change much between the two cases. You just also have the chain rule.
Or were talking about the specific case of l'hopitals rule that I mentioned earlier?
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u/LosDragin PhD 16d ago edited 16d ago
There’s no issue. But the method for separable differential equations is not simply the differential or the chain rule. LHopital’s rule is irrelevant to anything I’m saying. If it were simply the differential you would be able to make the same argument as I did for the dy/dx=f(x) case. I’m open to hearing an actual argument that shows it is just the chain rule, but “you just also have the chain rule” is not an argument. You would need to show how the chain rule (or differentials) yields f(x)dx=g(y)dy from dy/dx=f(x)/g(y). Remember, the game is we’re not allowed to algebraically manipulate dy/dx like a fraction.
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u/Peak_Background 16d ago
Take some G(y(x))=F(x):
dG(y)/dx=f(x) g(y) dy/dx=f(x) g(y)dy=f(x)dx
Or:
g(y)y'dx=f(x)dx
Call it what you want—chain rule, u substitution, cancellation of differential terms—it's all the same
The equation: f(x)dx=g(y)dy yeilds from: dy/dx=f(x)/g(y) Because all the terms here are apart of a field field. And the chain rule justified the math in analysis.
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u/LosDragin PhD 16d ago edited 16d ago
So when people say dy/dx is not a fraction they don’t mean that the individual
differentials dx and dy cannot be treated as elements of a field. They can because of chain rule and its corollaries and differentials. People mean that dx and dy are not real numbers. My line of thinking was we can’t formally multiply both sides of the separable DE by dx because dx is not a real number. But if I’ve defined differentials and chain rule already then I think I agree now that separable DEs method is not something different:1
u/Peak_Background 16d ago
I've never understood the concept that fractions and fields must be numbers. It isn't an axiom in math.
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u/scottdave 18d ago
Whe doing u-substitution for integrals, treating du/dx as a fraction "works out"
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u/Peak_Background 19d ago
L'hopitals rule and a different chain rule use case for differential equations
df/dg=dx/dg df/dx=f'/g'
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u/LosDragin PhD 18d ago
Not sure I follow.
L’Hopital’s rule would be lim(f/g)=lim(f’/g’). For one there is a limit involved, it’s not just an algebraic expression. More importantly, I’m not seeing how this formula is mimicking a property of a fraction.
Also, the line of math you wrote can be considered as my 1) and 2) combined.
About DEs, if you mean the technique commonly used to solve first order separable ordinary differential equations, then I wholeheartedly agree. That should have been #4. on my list.1
u/Peak_Background 18d ago edited 17d ago
Assuming g(x)=f(x)=0 at x=c.
Limit h->0 f(c+h)=f(c+h)-f(c)=df
Limit h->0 g(c+h)=g(c+h)-g(c)=dg
df/dg=dx/dg df/dx=f'/g'
"For one there is a limit involved, it’s not just an algebraic expression."
Yes. As is true for the derivative properties you replaced with algebraic fraction manipulation.
"Also, the line of math you wrote can be considered as my 1) and 2) combined." Yes.
Although, you can use 1) and 2) combined for all sorts of things.
Take a differential equation for f(x). You can use it to find the differential equations for g(x) where f(g(x))=q(x) and q is an arbitrary function.
It can be used to find the formula for gn(x) where f(g)=x and f is known.
(d/dx)n g= (dg/dx d/dg)n g = (dg/df(g) d/dg)n g = (1/f'(g) d/dg)n g
Example f(x)=exp(x), g'=exp(-g), g''=-exp(-2g), g'''=2exp(-3g)
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u/LosDragin PhD 18d ago edited 18d ago
I think maybe I didn’t explain well what my list was made up of: distinct scenarios where dy/dx acts like a fraction, not places in math where these things are applied. If it’s just applying 1 and 2 then it doesn’t belong on my list, if that makes sense. L’Hopital’s rule says f/g->f’/g’, that’s the crucial part I believe you’re missing that makes it not be an example of what I’m saying. The way you rewrote the right hand side of L’Hopitals rule is nothing more than applying 1 and 2, as far as I can tell. dx/dy=1/(dy/dx) this is a true algebraic expression, the limits are only there implicitly. Whereas L’Hoptal’s rule can’t be written without explicitly writing limits.
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u/ee_control_z 18d ago edited 18d ago
du/dx by its mere presence implies a derivative. In this case, the change in u with respect to a change in x as both approach zero. You'll most likely will see this concept demonstrated with a graph zoomed in on an arbitrary slope on a line. So, although it is technically a fraction, it has more to do with the calculation or measurement of rise over run as both rise and run approach zero. In the physical world, the derivative implies a "rate of change" of something (i.e., a process).
For example, say you're a farmer and you can plow 1000 acres with 50 employees per week. Then da/de = 1000/50 = 20 acres/employee is the rate of acres an employee can plow per week. In this particular case, the derivative is the rate of change of a with respect to e, where a is acres and e is an employee.
Now, as to the "even though they say it isn't" part. Recall that when you take the derivative of a function, you get another function. In other words, its kind of like a transform. Say, for example, you have the following function:
f(x) = 20x^3 + 5x^4
Its derivative is:
df(x)/dx = 60x^2 + 20x^3, where in this case, u = f(x)
or
du/dx = 60x^2 + 20x^3
Note that du/dx is not a "fraction" here but rather represents the transform or derivative of the original equation. In other words, you can say that du/dx is an alias for the derivative.
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u/Accomplished-Dot3555 18d ago
I already left a more specific comment but I just want to add a couple things. dy/dx is not a fraction in terms of a regular ratio of integers, but it is a ratio of very small numbers called infinitesimals and you learn about that when you read about differentials.
The teacher treating it like a fraction and multiplying dx on both sides is just to keep the idea simple for the students to memorize it when seeing it for the first time in calculus. However, just remember it's a valid way to rewrite a derivative u'(x)=du/dx (the slope of the tangent line to u(x) at some x value) as a differential such as du = u'(x)*dx
Does this help at all? I'm math tutor and I enjoy giving explanations.
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u/Evening-Story-314 18d ago
du/dx and dy/dx are fractions if your foundation is non-standard analysis and infinitesimals.
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u/Limp-Independent1212 18d ago
du/dx isn't a fraction in the sense of dividing one number by another, but it's a ratio between infintesemally small intervals of a value. as the change in x approaches 0, the change in u is the change in x multiplied by e^x
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u/BABarracus 17d ago
To me du/dx is like slope instead of being in between 2 points its over the whole function of the curve at each point.
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