Why Can Derivatives Be Treated Like Fractions in Solving Equations?

In summary, derivatives can be treated like fractions when it helps to understand the logic behind them. However, this is not always the case and there are other ways to think about derivatives.
  • #1
NoahsArk
Gold Member
237
22
TL;DR Summary
Not sure why derivatives are sometimes treated as fractions.
I don't understand the logic behind why derivatives can be treated like fractions in solving equations:

## \frac {du}{dx} = 2 ## simplified to
## du = 2dx ##

I keep seeing this done with the explanation that "even though ## \frac {du}{dx} ## is not a fraction, we can treat it like one". Why?

Thanks
 
Physics news on Phys.org
  • #2
Consider how dy/dx came about as the slope at any given point on a curve and that taught you start with ##\Delta y / \Delta x ## which a fraction of small differences.

Soon you will learn about differentials dy and dx and then it may make sense.
 
  • Like
Likes NoahsArk
  • #3
https://math.stackexchange.com/questions/1784671/when-can-we-not-treat-differentials-as-fractions-and-when-is-it-perfectly-ok#1784701 said:
So, yes, ##dy/dx## can be treated like a fraction in the sense (and to the extent) that the Chain Rule ##dy/dx=(dy/du)(du/dx)## is a thing that is true, but that's essentially as far as the fraction analogy goes. [...]

[...] here are examples of fraction-like manipulations which are not valid:
$$\left( \frac{dy}{dx} \right)^2 = \frac{(dy)^2}{(dx)^2} \ \ \text{ or } \ \ 2^{dy/dx} = \sqrt[dx]{2^{dy}}.$$
Because these manipulations are nonsensical, students are often warned not to treat derivatives like fractions.
 
  • Like
Likes NoahsArk and DaveE
  • #4
NoahsArk said:
Summary:: Not sure why derivatives are sometimes treated as fractions.

I don't understand the logic behind why derivatives can be treated like fractions in solving equations:

## \frac {du}{dx} = 2 ## simplified to
## du = 2dx ##

I keep seeing this done with the explanation that "even though ## \frac {du}{dx} ## is not a fraction, we can treat it like one". Why?

Thanks
Technically, in each case you have to prove (or at least justify) the manipulation. Although, the one you posted is usually taken as the definition of a differential.
 
  • Like
Likes NoahsArk
  • #5
Limits in Calculus often gave students headaches especially the epsilon definition.

in contrast, developing Calculus with hyperreals where the limit notion is buried inside the number system makes calculus easier and gives you a kind of differentials algebra where you can safely treat dy/dx as a fraction.

https://en.m.wikipedia.org/wiki/Hyperreal_number

one mathematician wrote a whole calculus book based on hyperreals. Limits were introduced later in the book for completeness.

https://people.math.wisc.edu/~keisler/calc.html
 
  • Skeptical
  • Like
Likes sysprog, Dale, weirdoguy and 2 others
  • #6
jedishrfu said:
in contrast, developing Calculus with hyperreals where the limit notion is buried inside the number system makes calculus easier and gives you a kind of differentials algebra where you can safely treat dy/dx as a fraction.
If a student is struggling with calculus, I'm not sure non-standard analysis is an easy escape route!
 
  • Like
Likes jasonRF, Delta2, DaveE and 1 other person
  • #7
Well if we are trying to integrate ## \frac {du}{dx} = 2 ## and we multiply both sides by dx:
du = 2dx
We end up with ## u\prime = 2 ## and then after integrating get u = 2x. So if the dx just disappears, what was the point of putting it on the right side? Why not just eliminate it from the equation? What does the dx even mean when it's being multiplied by 2?
 
  • #8
NoahsArk said:
Well if we are trying to integrate ## \frac {du}{dx} = 2 ## and we multiply both sides by dx:
du = 2dx
We end up with ## u\prime = 2 ## and then after integrating get u = 2x.
Actually, you get u = 2x + C, where C is an arbitrary constant.
NoahsArk said:
So if the dx just disappears, what was the point of putting it on the right side? Why not just eliminate it from the equation? What does the dx even mean when it's being multiplied by 2?
As far as integration is concerned, the dx (or dt or d<whatever>) indicates what the variable of integration is. Another explanation is that it's a holdover from the Riemann sum in which you add up a bunch of thin rectangles that are 2 units tall and ##\Delta x## wide. That is,
$$\sum_{i = 0}^n 2 \Delta x_i$$
Here, some interval [a, b] is subdivided into n subintervals, with the width of the i-th subinterval being ##\Delta x_i##. If you take the limit of the above sum, as ##n \to \infty##, you get the area below the line y = 2.

Actually, in this case, you get the same result for any positive integer n, but that doesn't happen for other functions, in general.
 
  • Like
Likes dextercioby, sysprog and NoahsArk
  • #9
jedishrfu said:
Limits in Calculus often gave students headaches especially the epsilon definition.

in contrast, developing Calculus with hyperreals where the limit notion is buried inside the number system makes calculus easier and gives you a kind of differentials algebra where you can safely treat dy/dx as a fraction.

https://en.m.wikipedia.org/wiki/Hyperreal_number

one mathematician wrote a whole calculus book based on hyperreals. Limits were introduced later in the book for completeness.

https://people.math.wisc.edu/~keisler/calc.html
I'm really glad you weren't my High School Calculus teacher, but then I soon learned at University (Group Theory, actually) that I'm more suited to Engineering than Mathematics.

Maybe that's something Calculus teachers should take into account. How many of their students will become Mathematicians and such, versus how many will need Calculus for practical uses.
 
  • #10
The way I learned it, is that I was being taught first the definition of the differential of a function f as $$df=f'(x)dx$$ and then the Leibniz symbolism for derivative as the ratio (fraction) of two differentials, the differential of the function to the differential of the identity function i(x)=x which is di=(x)'dx=dx. That is $$f'=\frac{df}{dx}=\frac{f'dx}{dx}=f'$$.
So , according to my opinion, it can be perfectly treated as a fraction. But beware if you have higher order derivatives for example $$f''=\frac{d^2f}{dx^2}$$ you can perfectly write this as $$d^2f=f''(x)(dx)^2$$ however you just can NOT integrate both sides of the last expression because the integral $$\int d^2f=\int f''(x)dx^2$$ is not well defined, we know that only integrals of the form $$\int g(x) dx$$ are well defined, that is we must have ##dx## there and not ##dx^2## or ##dx^n##.
 
  • Like
Likes NoahsArk
  • #11
NoahsArk said:
Well if we are trying to integrate ## \frac {du}{dx} = 2 ## and we multiply both sides by dx:
du = 2dx
We end up with ## u\prime = 2 ## and then after integrating get u = 2x. So if the dx just disappears, what was the point of putting it on the right side? Why not just eliminate it from the equation? What does the dx even mean when it's being multiplied by 2?
The simplest way to think about differentials is that they are like the smallest possible change in a quantity: ##dx## is like the smallest possible ##\Delta x##. From that point of view, you can multiply them by numbers.

To put the concept of a differential on a sound mathematical footing is not so easy, but that shouldn't be a concern for a first course in calculus.
 
  • Like
Likes Klystron
  • #12
DaveE said:
I'm really glad you weren't my High School Calculus teacher…

Why? The hyperreals are used merely to say it’s okay to treat them as fractions. Limits are then delayed until later chapters. Calculus then becomes more of applying the laws to functions to get derivatives and to solving realworld problems with it.

The point of the limit discussion early on is to allay feelings that the foundation of Calculus is somewhat sketchy. It’s why mathematicians worked so hard to come up with limit theory.

Look at the Keisler Calculus book and see if its alternative approach is good or not.
 
  • Like
Likes Dale
  • #13
NoahsArk said:
Well if we are trying to integrate ## \frac {du}{dx} = 2 ## and we multiply both sides by dx:
du = 2dx
We end up with ## u\prime = 2 ## and then after integrating get u = 2x.
You started with ##\frac{du}{dx} = u' = 2##. What you're really doing is
\begin{align*}
\frac{du}{dx} &= 2 \\
\int \frac{du}{dx}\,dx &= \int 2\,dx \\
u &= 2x + C
\end{align*} The cancelling of ##dx## that you often see is just a useful mnemonic that happens to work notationally, but as others have pointed out, if you try to take the analogy too far, you run into trouble.

So if the dx just disappears, what was the point of putting it on the right side? Why not just eliminate it from the equation? What does the dx even mean when it's being multiplied by 2?
Just as ##du/dx## really means applying the operator ##d/dx## to ##u##, you can think of the integration operator as not just ##\int## but ##\int dx##. You never just throw in an integral sign without differentials also appearing.
 
  • Like
Likes NoahsArk
  • #14
Maybe part of my confusion is coming from definitions. What's the difference between ## \frac {\Delta y}{\Delta x} ## and ## \frac {dy}{dx} ## ? I thought all these were just different ways of saying ## y\prime ## where it's assume ## y\prime ## means the amount that y changes for a given value of x.
 
  • #15
it is $$\frac{dy}{dx}=\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}$$ which intuitively means that the difference between the two get as small as we want, as long as we choose small enough ##\Delta x##.
 
  • Like
Likes bob012345, jack action, NoahsArk and 1 other person
  • #16
NoahsArk said:
## \frac {du}{dx} = 2 ## simplified to
## du = 2dx ##

I keep seeing this done with the explanation that "even though ## \frac {du}{dx} ## is not a fraction, we can treat it like one". Why?
They both say the same thing. It is a ratio between ##du## and ##dx##. Whatever the change in ##x## is, the change in ##u## is twice as large.
 
  • #17
FactChecker said:
They both say the same thing. It is a ratio between du and dx. Whatever the change in x is, the change in u is twice as large.
I think that's an easier way to think about it. So, ## \frac {du}{dx} ## really is a fraction in a way.
du = 2dx becomes ## u \prime = 2dx ##. I am assuming the "d" on the left side and the "dx" on the right then disappear in the next step since these are both operators and we are doing the inverse operation to end up with u = 2x + c?
 
  • Skeptical
Likes PeroK
  • #18
NoahsArk said:
I think that's an easier way to think about it. So, ## \frac {du}{dx} ## really is a fraction in a way.
du = 2dx becomes ## u \prime = 2dx ##. I am assuming the "d" on the left side and the "dx" on the right then disappear in the next step since these are both operators and we are doing the inverse operation to end up with u = 2x + c?
##du \ne u'##
 
  • Like
Likes vela
  • #19
@PeroK

I thought du, ## u \prime ##, and ## f \prime (x) ## all meant the same thing- i.e. the derivative of u with respect to x. What is the difference between du, ## u \prime ##? Thanks
 
  • #20
NoahsArk said:
@PeroK

I thought du, ## u \prime ##, and ## f \prime (x) ## all meant the same thing- i.e. the derivative of u with respect to x. What is the difference between du, ## u \prime ##? Thanks
There are two common notations for a derivative, namely ##f'(x)## and ##\frac{df}{dx}##. Or, if we write ##y = f(x)##, then ##\frac{dy}{dx} = f'(x)##.

The differentials ##dy## and ##dx## are not derivatives. It's not clear why you should think that ##dy = \frac{dy}{dx}##.
 
  • #21
But does ## u \prime = \frac {dy}{dx}? ##

I'm confused about what dy means standing by itself. ## \frac {dy}{dx} ## meant the derivative of y with respect to x. In the equation ## \frac {dy}{dx} = 2 ##, we simplified to dy = 2dx. Then in the next step (and please let me know if the next step I'm doing is wrong), we changed dy to ## u \prime ## and changed 2dx to just 2 then integrated. If the step was correct to change dy to ## u \prime ## and if ## u \prime ## is the same as ## \frac {dy}{dx} ## then we end up with the left side of the equation being the same with what we started with.

I am assuming the operator for the antiderivative is ## \int ##. Seems like it would be much easier to solve for ## \frac {dy}{dx} = 2 ## by saying ## \int \frac {dy}{dx} = \int 2 ## then we'd get y = 2x + c. This whole thing about putting the dx on the other side, then changing the dy to a ## y \prime ## which means the same thing as ## \frac {dy}{dx} ## which we started with, then erasing the dx from the right side which wasn't there to begin with, seems confusing.
 
  • #22
NoahsArk said:
But does ## u \prime = \frac {dy}{dx}? ##

I'm confused about what dy means standing by itself. ## \frac {dy}{dx} ## meant the derivative of y with respect to x. In the equation ## \frac {dy}{dx} = 2 ##, we simplified to dy = 2dx. Then in the next step (and please let me know if the next step I'm doing is wrong), we changed dy to ## u \prime ## and changed 2dx to just 2 then integrated. If the step was correct to change dy to ## u \prime ## and if ## u \prime ## is the same as ## \frac {dy}{dx} ## then we end up with the left side of the equation being the same with what we started with.

I am assuming the operator for the antiderivative is ## \int ##. Seems like it would be much easier to solve for ## \frac {dy}{dx} = 2 ## by saying ## \int \frac {dy}{dx} = \int 2 ## then we'd get y = 2x + c. This whole thing about putting the dx on the other side, then changing the dy to a ## y \prime ## which means the same thing as ## \frac {dy}{dx} ## which we started with, then erasing the dx from the right side which wasn't there to begin with, seems confusing.
It's not clear where you are getting this from. There is no need to introduce another variable ##u##, but if you do you have to define it.

##\int## means nothing on its own. It must always be ##\int dx## (or ##dy## or ##dt##), because you must always specific the variable with respect to which you are integrating. For example:
$$\frac{dy}{dx} = 2 \ \Rightarrow \ \int \frac{dy}{dx} dx = \int 2 dx = 2x + C$$Then we can use the fundamental theorem to see that $$\int \frac{dy}{dx} dx = y$$Note that we only need one arbitrary constant, so there's no point in adding another one here. In any case, this gives us $$y = 2x + C$$ More generally:
$$\frac{dy}{dx} = f(x) \ \Rightarrow \ \int \frac{dy}{dx} dx = \int f(x) dx \ \Rightarrow \ y = \int f(x) dx$$
 
  • Like
Likes NoahsArk and vela
  • #23
##dy## represents the change in y, not the ratio of change in y divided by change in x. In general use, ##dy## represents a minuscule change in y although it might be a larger change in some contexts.
 
  • Like
Likes NoahsArk
  • #24
Mathematicians timidly say they are not fraction. They are right, maybe. I observe many physics people including myself treat them as if they are fractions and have not encountered serious problems as far as I am concerned.
 
  • Like
Likes NoahsArk
  • #25
jedishrfu said:
Limits in Calculus often gave students headaches especially the epsilon definition.

in contrast, developing Calculus with hyperreals where the limit notion is buried inside the number system makes calculus easier and gives you a kind of differentials algebra where you can safely treat dy/dx as a fraction.

https://en.m.wikipedia.org/wiki/Hyperreal_number

one mathematician wrote a whole calculus book based on hyperreals. Limits were introduced later in the book for completeness.

https://people.math.wisc.edu/~keisler/calc.html
I have always liked the hyperreal approach and I wonder why calculus curriculums have been so slow to pick it up. It is not exactly a new thing any more. It seems far more intuitive and natural than the whole epsilon delta limits approach.
 
Last edited:
  • Like
Likes jedishrfu
  • #26
NoahsArk said:
@PeroK

I thought du, ## u \prime ##, and ## f \prime (x) ## all meant the same thing- i.e. the derivative of u with respect to x. What is the difference between du, ## u \prime ##? Thanks
No. du is the (possibly infinitesimal) change in u. Period.
##u\prime## is the ratio of change in u divided by the associated change in x.
It is like the difference between 55 and 55/3. They are not the same.
 
  • #27
Dale said:
I have always liked the hyperreal approach and I wonder why calculus curriculums have been so slow to pick it up. It is not exactly a new thing any more. It seems far more intuitive and natural than the whole epsilon delta limits approach.
What would be your definition of the limit of a convergent sequence?
 
  • #28
PeroK said:
What would be your definition of the limit of a convergent sequence?
For teaching calculus I would skip the construction of hyperreals in terms of convergent sequences. That is not necessary for calculus students. I would focus on the properties of hyperreals and how to use them rather than how to construct them.
 
  • #29
Dale said:
For teaching calculus I would skip the construction of hyperreals in terms of convergent sequences. That is not necessary for calculus students. I would focus on the properties of hyperreals and how to use them rather than how to construct them.
You're proposing not to teach sequences at all? Go straight to hyperreals?

You wouldn't teach anything to do with standard real analysis?
 
  • #30
PeroK said:
You're proposing not to teach sequences at all? Go straight to hyperreals?

You wouldn't teach anything to do with standard real analysis?
In a calculus class I would teach only as much real analysis or sequences as is absolutely necessary to teach calculus. I think that is pretty minimal, but "not at all" and "wouldn't teach anything" probably goes a bit too far. The goal is calculus, not these other topics.
 
  • #31
Dale said:
In a calculus class I would teach only as much real analysis or sequences as is absolutely necessary to teach calculus. I think that is pretty minimal...
I think that would not be wise. Engineers and applied mathematicians would probably be hurt by that.
 
  • #32
FactChecker said:
I think that would not be wise. Engineers and applied mathematicians would probably be hurt by that.
How?
 
  • #33
Dale said:
How?
Engineers and applied mathematicians need to be very experienced and comfortable with convergent sequences and series of all sorts.
 

1. Why can derivatives be treated like fractions in solving equations?

Derivatives can be treated like fractions in solving equations because they represent the rate of change of a function at a specific point. This means that they can be manipulated using the same rules as fractions, such as the quotient rule and chain rule, to solve equations.

2. What are the benefits of treating derivatives like fractions?

Treating derivatives like fractions allows us to use familiar algebraic techniques to solve equations involving derivatives. It also makes it easier to find the maximum and minimum values of a function, as well as the points where the function is increasing or decreasing.

3. Are there any limitations to treating derivatives like fractions?

Yes, there are limitations to treating derivatives like fractions. This method may not work for all types of equations, such as those involving logarithmic or trigonometric functions. It also may not be applicable when dealing with higher order derivatives.

4. How does treating derivatives like fractions relate to the concept of the chain rule?

The chain rule is a method used to find the derivative of a composite function. When treating derivatives like fractions, the chain rule can be applied to manipulate the derivatives in a similar way as fractions, allowing us to solve equations involving composite functions.

5. Can derivatives be treated like fractions in all types of mathematical problems?

No, derivatives cannot be treated like fractions in all types of mathematical problems. This method is specifically applicable to solving equations involving derivatives. In other types of problems, such as finding the area under a curve, derivatives cannot be treated like fractions.

Similar threads

Replies
22
Views
2K
Replies
6
Views
2K
Replies
4
Views
287
Replies
4
Views
3K
  • Calculus
Replies
15
Views
1K
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
754
  • Calculus
Replies
2
Views
1K
Replies
2
Views
1K
Replies
6
Views
2K
Back
Top