Discussion on $\frac{d}{dx}$

It is a common practice to move the $dx$ around when solving ODE and we take for granted when we integrate both sides. However, I've been rather uncomfortable with this. From an analysis perspective, $dx$ itself doesn't make sense to me. Instead, we always consider $\frac{df(x)}{dx}$ , or the operator $\frac{d}{dx}$ by itself.

Like if we consider the example $\frac{dy}{dx}$ = $e^x$ . Here to solve this equation we take $dx$ to right hand side and then integrate. But $\frac{d}{dx}$ is considered as an operator.So why we treat it as a fraction in the above example .

Kindly explain me the process behind this or if there is any kind of theory.

If I mentioned something wrong (concept or anything), kindly also mention it.

#Calculus

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

$df(x):=\lim_{\triangle x \to 0}[f(x+\triangle x)-f(x)]$ Using this we can find that : $\dfrac{df(x)}{dx}=\dfrac{\lim_{\triangle x \to 0}[f(x+\triangle x)-f(x)]}{\lim_{\triangle x \to 0}[\cancel{x}+\triangle x-\cancel{x}]}$ $=\lim_{\triangle x \to 0}\dfrac{f(x+\triangle x)-f(x)}{\triangle x}$

Zakir Husain - 3 months, 2 weeks ago

I have always considered $\text{d}x$ to be an infinitesimal and $\text{δ}x$ as an infinitesimal in a given direction, so moving them around always made sense to me.

Jason Gomez - 3 months, 3 weeks ago

Can you please elaborate a bit? I am not getting you that much clearly.

Rajyawardhan Singh - 3 months, 2 weeks ago

Ok so my definition of $\frac{\text dy}{\text dx}$ in my head is always $\displaystyle \frac{\displaystyle\lim_{Δy ➝ 0} Δy}{\displaystyle\lim_{Δx ➝ 0} Δx}$ , so I can easily multiply $\text dx$ anywhere I want without guilt, having this definition in mind

Jason Gomez - 3 months, 2 weeks ago

In otherwords, I don’t see it as an operator, but an actual division\fraction of two numbers(they are not called numbers, but can be vaguely described in terms of them)

Jason Gomez - 3 months, 2 weeks ago

@Jason Gomez – Thanks man . Your thinking is nice . Its really a good approch to understand it.

Rajyawardhan Singh - 3 months, 2 weeks ago

Correct, I have the same concept in mind. "The symbol Δ refers to a finite variation or change of a quantity – by finite, I mean one that is not infinitely small. The symbols d,δ refer to infinitesimal variations or numerators and denominators of derivatives. "

Mahdi Raza - 2 months, 1 week ago

Kindly look in my calculus notes. in chapter 2, it is mentioned that $dy$ stands for change in y-coords as $dy$ approaches 0, same for $dx$ . Multiplying a fraction by its denominator is what we often do, right?

Jeff Giff - 3 months, 2 weeks ago

So, having $\dfrac{dy}{dx}$ a fraction instead of an operator, things get easier :)

Jeff Giff - 3 months, 2 weeks ago

Thanks everyone these really helped! I was also thinking of a different approch. Like we consider some new $Dx$

And multiplying it to both sides like

$\frac{dy}{dx}$ $Dx$ = $e^x$ $Dx$

Now integrating both sides vanishes the operator .

If only we can get $Dx$ in terms of $dx$ ,

We are done!

Rajyawardhan Singh - 3 months, 2 weeks ago

Why not you just put $Dx$ equal to $dx$

Jason Gomez - 3 months, 1 week ago

I was just taking some arbitary $Dx$ and tried to show it equal to $dx$ .

Rajyawardhan Singh - 3 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Discussion on ddx\frac{d}{dx}dxd​

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Discussion on $\frac{d}{dx}$