# Why Do People Treat dy/dx as a Fraction?

#### (New question of the week)

Here is a recent question from Fida, another long-time “patient” of ours at Ask Dr. Math:

I hope you won’t mind answering a question.

I have seen people treating dy/dx as  fraction although it is not. I am guessing it is a handy shortcut. But what is the proof for that?

e.g. in U-substitution

Here is a fragment where they do this trick

u=sin x
du/dx = cos x
du = cos x dx

In my response, I quoted an unpublished question and answer of  mine from 2015, which in turn refers to several related answers in the archive:

What you say is exactly how I think of it: the notation is a shortcut to the application of various theorems. I would not say there is a single proof for this in general, but rather each theorem (e.g. integration by substitution) is proved without using the fraction notation, and then we note that the content of the theorem essentially says that you can, in this setting, treat dx and dy as individual entities in a certain way.

The underlying reason, ultimately, is that the derivative is defined as a limit of fractions, and so properties of fractions “survive the passage to the limit”.

Question:
So recently we've been going over integration in Calculus. In subjects like integration by substitution and differential equations she said that you had to cross multiply dy/dx to isolate either dy or dx for integration. I always thought dy/dx was a function like a sine or cosine and I don't see why you can break up dy/dx by cross multiplying. What do dy and dx represent?

I've always thought dy/dx represented a type of function like logs or trig functions. Obviously, you can't break up sin(x) into s * in(x). I thought the same rule applied to dy/dx. I don't really understand what dy and dx are individually and why when divided, you get the derivative of a function.

Integration by substitution:

INT[sin(x^2)2x dx]

Let u=x^2
du/dx=2
2x dx =du  The Cross-Multiplying step I don't understand
dx= du/2x

INT[sin(x^2)2x dx]
INT[sin(u)2x  du/2x)
INT[sin(u) du]
-cos(u)
-cos(x^2)

Differential Equation:

dy/dx = 2x/y
y dy = 2x dx  The Cross-Multiplying step I don't understand
INT[y dy] = INT[2x dx]
y^2/2 = x^2 + C
y^2 = 2(x^2) + C
y = sqrt(2(x^2) + C)

I understand the thought processes behind substitution and differential equations. I have no problem with doing the homework problems. I just have a problem understanding what dy and dx are individually why dividing them gets the derivative of a function.

I like your illustration of s*in(x)! You're right that dy/dx is defined as a single entity, f'(x), and is not really the fraction it looks like. But ...

Here are some explanations of aspects of what is going on here:

Differentials
http://mathforum.org/library/drmath/view/53678.html

Can dy/dx Be Treated as a Fraction?
http://mathforum.org/library/drmath/view/65462.html

Why Does Integration by Substitution Work?
http://mathforum.org/library/drmath/view/69805.html

Differential of a function
http://en.wikipedia.org/wiki/Differential_of_a_function

In one sense, you can just think of it all as a notation that works: because of several theorems that can be proved using the definition of the derivative, it turns out that (in specified situations) you can treat dy/dx and the integral AS IF dy/dx were a fraction and the dx in the integral were its denominator, even though they really are not. The things you are doing with "fractions" are just ways to remember the results of those theorems. The reason this works out involves the fact that the derivative is defined as the LIMIT of a fraction, and much of its behavior still looks like a fraction; and similarly that the definite integral is defined as the limit of a sum of products with delta x, and its behavior still reflects that (within reason). In another sense, you can define something called a differential (dx, dy) in such a way that the derivative really IS the ratio of these. This can be done in a couple different ways, depending the level of rigor you want in your proofs (and the difficulty you can handle in understanding what they are talking about). Take a look at the pages I referred to, and we can discuss any questions you still have!

Fida, you can find proofs of integration by substitution by searching for that phrase; here is one good source: