# Why is Delta y not equal to dy for infinitesimally dx on the graph?

• B
• Mike_bb
In summary: No, I wouldn't. If someone wants to use d, they should be willing to learn it properly. No, I wouldn't. If someone wants to use d, they should be willing to learn it properly.
Mike_bb
TL;DR Summary
Why is Delta y not equal to dy for infinitesimally dx on the graph?
Hello!

As is known, $\Delta y = dy$ for infinitesimally small $dx$. It's true.
But if we have graph we may see that $\Delta y$ isn't equal to $dy$ even for infinitesimally small $dx$. Why is that so?

Thanks!

Last edited:
Please forget the notation ##dy## and ##dx## at this point in your education. They are no actual distances and are only an abbreviation in this context.

Firstly, they make only sense in the combination ##\dfrac{dy}{dx}## in this case.
Secondly, they only abbreviate a limit: ##\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}.##

sysprog
Mike_bb said:
Summary:: Why is Delta y not equal to dy for infinitesimally dx on the graph?

Hello!

As is known, $\Delta y = dy$ for infinitesimally small $dx$. It's true.
But if we have graph we may see that $\Delta y$ isn't equal to $dy$ even for infinitesimally small $dx$. Why is that so?

Thanks!View attachment 301195
Because delta y is the change in y, while dy is only the linear part.

martinbn said:
Because delta y is the change in y, while dy is only the linear part.
Yes. But we can use dy to represent finite Delta y as sum of infinitesimally small dy and it's right.

PeroK
Mike_bb said:
That is exactly why I said: forget about it.

It is a misleading notation until you are willing to learn it rigorously, in which case it becomes complicated. As long as you have to use the term "slope", as long are the actual quantities
$$\Delta x\, , \,\Delta y\, , \,\Delta f(x)\, , \,\dfrac{\Delta y}{\Delta x}\, , \,\lim_{\Delta x \to 0} \dfrac{\Delta y}{ \Delta x}$$
all you actually need. Write it ##y'## and ##f'(x).##

Or learn it correctly. In this case, you should start to read the series I linked to. However, the first part is all you need at this point. Parts 2-5 show you where such a notation leads to.

fresh_42 said:
Please forget the notation ##dy## and ##dx## at this point in your education. They are no actual distances and are only an abbreviation in this context.

Firstly, they make only sense in the combination ##\dfrac{dy}{dx}## in this case.
Secondly, they only abbreviate a limit: ##\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}.##
Why would you want to forget about d it is often used.
$$\frac{\mathit{dy}}{\mathit{dx}}=\frac{\mathit{dx}}{\mathit{du}}\frac{\mathit{du}}{\mathit{dx}}$$
$$\int \frac{\mathit{dx}}{a x^2+b x+c}$$
$$u \mathit{dx}=u v-v \mathit{dx}$$
Some people write
$$\int x^2 \phantom{\mathit{dx}}=\frac{1}{3}x^3+C$$
but they know d is there without writing it like
$$x=1x^1$$.
It is not clear that $$y^\prime$$ is a superor notation but regardless one must know all the commonly used notations to avoid confusion.
Would you tell someone to forget $$\sin$$ and write
$$\mathrm{sen}\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$$?

lurflurf said:
Why would you want to forget about d it is often used.
Because it confuses you and creates more problems than it solves. The chain rule can easily be phrased with primes instead of differentials, and the expression in an integral is only there to mark the variable anyway.

If you want to use it, then please explain it to me. I know that my definition won't be the one you understand (https://en.wikipedia.org/wiki/One-form; https://www.physicsforums.com/threa...ifferential-forms.1012875/page-2#post-6630386), so give me yours. What are ##df## and ##dx##?

lurflurf said:
$$\frac{\mathit{dy}}{\mathit{dx}}=\frac{\mathit{dx}}{\mathit{du}}\frac{\mathit{du}}{\mathit{dx}}$$
$$\int \frac{\mathit{dx}}{a x^2+b x+c}$$
$$u \mathit{dx}=u v-v \mathit{dx}$$
Some people write
$$\int x^2 \phantom{\mathit{dx}}=\frac{1}{3}x^3+C$$
but they know d is there without writing it like
$$x=1x^1$$.
It is not clear that $$y^\prime$$ is a superor notation but regardless one must know all the commonly used notations to avoid confusion.
Would you tell someone to forget $$\sin$$ and write
$$\mathrm{sen}\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$$?

Last edited:
robphy

## 1. Why is delta y not equal to dy on the graph?

Delta y and dy represent two different concepts in mathematics. Delta y is the change in the y-coordinate of a point on a graph, while dy represents the derivative of a function at a specific point. These two concepts are not equivalent and cannot be used interchangeably.

## 2. Can delta y and dy be equal for infinitesimally small values of dx?

No, delta y and dy will still represent different values even for infinitesimally small values of dx. This is because delta y represents the change in the y-coordinate over a finite interval, while dy represents the slope of the tangent line at a specific point on the graph.

## 3. Why is it important to understand the difference between delta y and dy?

Understanding the difference between delta y and dy is crucial in calculus and other mathematical fields. It allows us to accurately calculate and interpret derivatives, which are essential in solving many real-world problems. Confusing delta y and dy can lead to incorrect results and misunderstandings.

## 4. Can delta y and dy ever be equal?

Delta y and dy can be equal in certain situations, such as when the function is a straight line. In this case, the slope of the tangent line (dy) will be equal to the change in the y-coordinate (delta y). However, this is not always the case and should not be assumed to be true for all functions.

## 5. How do we calculate delta y and dy on a graph?

To calculate delta y, we need to find the difference between the y-coordinates of two points on the graph. To calculate dy, we need to find the derivative of the function at a specific point, which can be done using various methods such as the limit definition or rules of differentiation.

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