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

Question

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?

Answer 1

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

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:

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

Answer 2

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}