Discussion on ddx\frac{d}{dx}

It is a common practice to move the dxdx 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, dxdx itself doesn't make sense to me. Instead, we always consider df(x)dx\frac{df(x)}{dx}, or the operator ddx\frac{d}{dx} by itself.

Like if we consider the example dydx\frac{dy}{dx} = exe^x. Here to solve this equation we take dxdx to right hand side and then integrate. But ddx\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.

Note by Rajyawardhan Singh
5 months, 1 week ago

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df(x):=limx0[f(x+x)f(x)]df(x):=\lim_{\triangle x \to 0}[f(x+\triangle x)-f(x)] Using this we can find that : df(x)dx=limx0[f(x+x)f(x)]limx0[x+xx]\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}]} =limx0f(x+x)f(x)x=\lim_{\triangle x \to 0}\dfrac{f(x+\triangle x)-f(x)}{\triangle x}

Zakir Husain - 5 months ago

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

Jason Gomez - 5 months, 1 week ago

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

Rajyawardhan Singh - 5 months, 1 week ago

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Ok so my definition of dydx\frac{\text dy}{\text dx} in my head is always limΔy0ΔylimΔx0Δx\displaystyle \frac{\displaystyle\lim_{Δy ➝ 0} Δy}{\displaystyle\lim_{Δx ➝ 0} Δx}, so I can easily multiply dx\text dx anywhere I want without guilt, having this definition in mind

Jason Gomez - 5 months, 1 week ago

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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 - 5 months, 1 week ago

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

Rajyawardhan Singh - 5 months, 1 week ago

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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 - 3 months, 4 weeks ago

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

Jeff Giff - 5 months ago

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

Jeff Giff - 5 months ago

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

And multiplying it to both sides like

dydx\frac{dy}{dx}DxDx = exe^xDxDx

Now integrating both sides vanishes the operator .

If only we can get DxDx in terms of dxdx ,

We are done!

Rajyawardhan Singh - 5 months ago

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Why not you just putDxDx equal to dxdx

Jason Gomez - 5 months ago

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I was just taking some arbitary DxDx and tried to show it equal to dxdx.

Rajyawardhan Singh - 4 months, 4 weeks ago

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