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In the dx/dy notation, we could interpret it is a fraction, right? Thus, the Ds cancel out? So that would result in x/y? Am I right?

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Professional Coder
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    It's meant to be suggestive of a fraction but it's not really. Don't treat them as fractions or you'll find into problems. – CyclotomicField Apr 29 '23 at 13:55
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 29 '23 at 14:31
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    It signifies a limit of ratios of differences, hence the d. You are essentially trying to cancel out the word 'difference'. – Paul Apr 29 '23 at 14:54
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    No, $dx$ is not a product $d$ times $x$. In Europe (but not the US), they may show this typographically by writing $\mathrm{d}x$ where the $d$ is in roman type to show it is an "operator". – GEdgar Apr 29 '23 at 14:59
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    Assuming you are not trolling, yes, it is possible to interpret $\mathrm{d}x/\mathrm{d}y$ as a fraction with some limitations. But, it is the ratio of $\mathrm{d}y$ and $\mathrm{d}x$, where $\mathrm{d}$ itself is not a variable. So, you can't cancel out $\mathrm{d}$. It is like cancelling $1$s in the $a_1/b_1$. – ck1987pd Apr 29 '23 at 15:06
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    $\require{cancel} \frac{\sin x}{n} = \frac{\operatorname{si!\cancel{n}} x}{\cancel n} = \operatorname{si}x = 6.$ – Misha Lavrov Apr 29 '23 at 15:30

2 Answers2

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You cannot, but you can separate dx from dy but you can't separate the ds. What the notation is saying is Derivative of x with respect to y, if you take away the ds the whole meaning changes.

Golden Boy
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    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Apr 29 '23 at 15:23
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No that is not correct. In ${df}\over{dt}$ notation, $df=f'(t)dt$, where $f'(t)$ is derivative of $f(t)$. Both $df$ and $dt$ represent infinitely small difference in $f$ and $t$ respectively. $dt= \lim_{Δt\rightarrow 0}Δt$. Working out the definitions ${{df}\over{dt}} =f'(t) = {\lim_{Δt\rightarrow 0}{{f(t+Δt)-f(t)}\over{Δt}}}$ is simply the derivative of $f$, where $Δt$ is a quantity not equal to zero considered as a difference between two values of $t$, but which can be taken to be as small value as desired (but not zero, since that causes divide by zero). For more info, see wikipedia entry on Derivative, https://en.wikipedia.org/wiki/Derivative?wprov=sfla1

Esa Pulkkinen
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