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Is the following true?

$$\frac{dy}{dx} = \frac{d(y+c)}{dx}$$

where $c$ is an arbitrary real constant.

I believe it is true, and my reasoning goes like this:

$dy$ is an infinitesimal, so the addition of another constant would still be an infinitesimal. I do not know if my reasoning is correct.

Do note that I'm not familiar with epilson delta and university calculus. I would appreciate it if someone could explain the above simply.

EDIT: I couldn't see why $d(y+c)= dy+dc$. What is $d$? Is it a number, or a function?

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  • $\begingroup$ You conclusion is correct. As you are reasoning by "non-standard" analysis, it is tough to call it "correct" but your intuition is leading you to the right places. $\endgroup$ – Doug M Apr 7 '17 at 16:24
  • $\begingroup$ @DougM Inasmuch as the OP is unfamiliar with "university calculus," it is a safe bet that the OP is equally unfamiliar with non-standard analysis. ;-)) $\endgroup$ – Mark Viola Apr 7 '17 at 16:26
  • $\begingroup$ I have some knowledge about the difference between standard and non-standard calculus analysis. It would be great if someone could explain this through simple, non-standard perspective. $\endgroup$ – Kyoma Apr 7 '17 at 16:29
  • $\begingroup$ How about this... differentiation is a "linear operation" that is $\frac {d}{dx} (f(x) + g(x)) = \frac {df}{dx} + \frac {dg}{dx}$ and since $c$ is constant $\frac {d}{dx} (y + c) = \frac {dy}{dx}$ $\endgroup$ – Doug M Apr 7 '17 at 16:31
  • $\begingroup$ I'm also just learning about the non-standard approach, but I think the story is the same in the standard and non-standard approaches. Since $c$ is constant, an infinitesimal change in the "argument" of $c$ changes nothing. If $c$ were a function, then we would argue that $d(y+c) = dy + dc$, since an infinitesimal change in the argument produces an infinitesimal change in both terms. $\endgroup$ – Alfred Yerger Apr 7 '17 at 16:34
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$$\mathrm{d}(y+c) = \mathrm{d}y + \mathrm{d} c $$

However, if $c$ is a constant, then

$$ \mathrm{d} c = 0 $$

so we get

$$\mathrm{d}(y+c) = \mathrm{d}y$$

Consequently, if one side of the following makes sense, then both sides do and they are equal:

$$\frac{\mathrm{d}(y+c)}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}x}$$

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  • $\begingroup$ I couldn't see why $d(y + c) = dy + dc$. What is $d$? Is it a number, or a function? $\endgroup$ – Kyoma Apr 8 '17 at 8:20
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let $g(x) = y(x) + c\;$ $ \forall x \in D_y$

\begin{align} & \frac{d(y+c)}{dx} = \frac{dg}{dx} = g'(x)= \lim_{h\to0} \frac{g(x+h)-g(x)}{h} =\lim_{h\to0} \frac{y(x+h)+c-y(x)-c}{h} \\[10pt] = {} & \lim_{h\to0} \frac{y(x+h)-y(x)}{h} = y'(x) = \frac{dy}{dx} \end{align}

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