2

When I was learning implicit differentiation in my class I was told to not think of $dy/dx$ as a fraction. Now I am doing integration by $u$-substitution and we treat $du/dx$ as a fraction and solve for $dx$. Why is this?

Michael Hardy
  • 1
SIDDHARTH SALAPAKA
  • 161
  • 1
  • 4
  • 1
    In my opinion, the answer to your question is not short. You want to find a Calculus textbook (like the one by Apostol) that mixes problem solving with (Real Analysis) proofs. Then find the portion in that book that discusses integration by substitution. Then, if needed, work backwards, until you nail down why informally treating $(du/dx)$ as a fraction is valid. – user2661923 Jan 22 '21 at 03:07
  • 1
    Related: https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio?noredirect=1&lq=1 – FoiledIt24 Jan 22 '21 at 03:09
  • to really understand this notion you have to study differential forms – user29418 Jan 22 '21 at 03:25
  • We've talked about this often. Short version is that there is disagreement as to whether or not it is a fraction. People who don't like infinitesimals say that it isn't. People who like infinitesimals (myself included) view it fundamentally as a fraction. As far as I'm aware there is no contradiction in using it as a fraction. However, for second and higher derivatives, you need a non-standard notation to keep using it as a fraction (i.e., $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$, which you can get by applying the quotient rule to the first derivative). – johnnyb Jan 22 '21 at 04:40
  • See my paper, "Extending the Algebraic Manipulability of Differentials" for more information. – johnnyb Jan 22 '21 at 04:40

1 Answers1

0

In the 19th century mathematicians became extremely squeamish about anything that is not logically rigorous, and they toss around the word "intuitive" promiscuously without noticing that that term admits a variety of different meanings to refer to non-rigorous things.

In many contexts, thinking of $dx$ as an infinitely small increment of $x$ and $dy$ as the corresponding infinitely small increment of $y$ is very useful.

Some instances of treating $dy/dx$ as if it were just an ordinary fraction can be justified by the chain rule, and among those are the things done in $u$ substitution.

Michael Hardy
  • 1