Is dy/dx An Operator or Symbol

[Ted_Grendy]

Hello all

My apologies for asking questions that maybe seen as simple, but I have been struggling with this for a while and every time I do some digging I get more confused. My question is this: -

Is dy/dx a symbol/operator that cannot be split

Or

Is dy/dx a fraction.

The reason why I am confused is because I was taught that to find the derivative of a function (which is to determine the slope at a given x value) we do the following: -

lim h ->0 [f(x+h) – f(x)] / h​

All of this is equal to dy/dx so for me when I see dy/dx I immediately think dy/dx is an operator just like + or – or *.

If I were to apply this to an actual function such as x^2 I would get 2x which would give me a real finite value for a given x.

The confusion is that I have now been introduced to differentials which splits dy/dx to something like: -

dy = f’(x).dx​

And now I am told that dy and dx are just very small changes in x and y.

I am struggling to get my head around how and why we can split dy/dx if dy/dx is equal to him h ->0 [f(x+h) – f(x)] / h – this doesn’t make any sense to me.

Can anyone help?

Thank you.

 

[Otis]

dy/dx can be a single symbol that represents the first derivative of a function named y, in terms of x. With this interpretation, we may not treat dy or dx as numbers.

dy/dx can be a ratio of the differentials dy and dx. With this interpretation, dy and dx each represent very small numbers.

The interpretation depends on the context. I've seen calculus courses that introduce derivatives using differentials; some of these courses never mention limits.

When I studied introductory calculus, the professor stressed that dy/dx is NOT a fraction, and he said, "Don't treat it like that." A few weeks later, he began treating dy/dx as a fraction. It was confusing at first, but with practice both notations eventually made sense.

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EDIT: Added last paragraph

 

[Otis]

Here are two videos. Feel free to skip ahead to [00:53], in the first video. (You won't miss anything pertinent.) ?

 

[Ted_Grendy]

Hi Otis

I have to admit I'm still struggling. How about this, given the function:-

f(x) = x^2+x

dy/dx = 2x+1

if I were to multiply both sides by dx I would get:-

dy = (2x+1)*dx

Then what exactly am I doing here what does this function tell me?

Thanks

 

[Dr.Peterson]

First, to answer your question about operators, "d/dx" can be thought of as an operator that converts a function f(x), or y, to its derivative, the function dy/dx or d/dx f(x). It can also be represented by " ' ", which converts function f to its derivative, the function f'.

Now, there are several ways to think about differentials (which is what we call dx and dy), but one that fits your question here is that dx is a small change in x, and dy = (2x+1)*dx is the resulting small change in y when you replace f(x) with its tangent line at (x, y). That is, you are multiplying a "run" by the "slope" to find the "rise" along the straight line.

In other contexts, you can think of it as just a "formal" notation that allows you to write integrals or differential equations neatly, which happens to work because of theorems that prove it does (not merely because it feels natural). In particular, it works ultimately because the derivative dy/dx is a limit of a fraction, ∆y/∆x, and some properties of the fraction "survive the passage to the limit".

 

[tkhunny]

Simple answer: Yes

No need to struggle. Different purposes at different times.

Similar Example from Physics: Is light particle or wave? Yes. Which one does your theory need?

Generally, notation should HELP you. Don't get hung up on it. Here's an interesting discussion:

Make sure YOU understand it and you can communicate what you understand.

 

[Otis]

… f(x) = x^2+x … dy/dx = 2x+1

if I were to multiply both sides by dx I would get: dy = (2x+1)*dx

… what does this [last equation] tell me?

It tells you that multiplying the rate at which y changes by a small change in x yields the corresponding small change in y.

For comparison, you know: Distance = Rate × Time. Let's say the rate is 4mph. If we let x represent elapsed-time and y represent distance-traveled, then we can write the function:

y = 4x

If x changes by 2 hours, then y changes by 8 miles.

Let's change x by a very small amount, instead. If x changes by 0.0000001 hours, then y changes by 0.0000004 miles. We see that a small change in x leads to a small change in y.

We've calculated changes in y by multiplying a change in x by the rate at which y changes with respect to x.

---

The same is true of your example function: y=x^2+x

You found the rate at which y changes: 2x+1

Using differentials, if we multiply a small change in x by the rate at which y changes, we will get the corresponding (small) change in y:

dy = (2x+1) ∙ dx

A small change in x multiplied by the rate at which y changes yields the corresponding small change in y.

This works with all continuous functions because functions behave linearly at infinitesimally-small scales. Use a computer to graph your function y=x^2+x. That polynomial graph is not a straight line. Now, zoom in far enough on a section (such that the change in x is very small). You will see that the function behaves linearly over very small changes in x and y.

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[Steven G]

Maybe this will help.
\(\displaystyle \frac{d}{dx}\) is an operator just like + and -. \(\displaystyle \frac{d}{dx}\) tell you to "take the derivative". For example \(\displaystyle \frac{d}{dx}\)(x^2+ 2x) means to take the derivative of x^2 + 2x. Now if the function is simply y, you can write \(\displaystyle \frac{d}{dx}\)(y) or \(\displaystyle \frac{dy}{dx}\) to denote to take the derivative of y. You can even write \(\displaystyle \frac{d(x^2+ 2x)}{dx}\) for \(\displaystyle \frac{d}{dx}\)(x^2+ 2x)