Calculus: Why do we ignore dx? [duplicate]

I'm new to calculus and I find it really confusing why we just ignore the dx at the end.
For example, when working on derivation of $x^2$, at the last step, we're left with

$f'(x)= 2x + dx$

But I've heard people in videos say: "Since $dx$ is super super super small, we can safely ignore it. But just ignoring it bugs me. If we choose to ignore it, should it not actually look like this:

$f'(x) \approx 2x $

Thanks for your time. Cheers :)

The definition of $f'(x)$ involves taking a limit:

$$ f'(x) = \lim_{\Delta x\to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}. $$

For $f(x)=x^2$, you end up with $$f'(x)=\lim_{\Delta x\to 0} (2x + \Delta x)$$ and in the limit the second summand vanishes. It is not ignored, it becomes zero in the limit, which the derivative is:

$$f'(x)=\lim_{\Delta x\to 0} (2x + \Delta x) = 2x + 0 = 2x.$$

All the "$=$"-signs in this equation are actual equalities, there are no approximations done here.

$\newcommand{\D}{\mathit{\Delta}}$

You are starting from $$ \frac{\D y}{\D x}=\frac{f(x+\D x)-f(x)}{\D x}=2x+\D x $$ The derivative is the limit for $\D x\to 0$, where only $f'(x)=2x$ remains. In other words, the derivative is the slope of the tangent to $f$ in $x$.

You have to make clear when you work with infinitesimals and when with symbolic expressions. Then in non-standard terms, the derivative is the standard part of the difference quotient, that is, from $f'(x)\approx 2x+dx$ with $dx\approx 0$ it follows again that $f'(x)=2x$.

It boils down to the precise meaning of $ f'(x) $, which may go beyond the stage you have reached so far.

Strictly $f'(x)$ is the limit of $\big(f(x+\delta x)-f(x)\big)/\delta x$ as $\delta x$ approaches zero. There is also a strict definition of what limit means: given any expression $E(\delta x) $ with $\delta x$ in it, it has the limit $\ell$ as $\delta x$ tends to zero if for every small positive number $\epsilon$ you can make $$ |E(\delta x) - \ell| \leqslant \epsilon $$ for all sufficiently small but not zero $\delta x$.

Written even more precisely, if for every $\epsilon >0$ there exists a number $\Delta > 0$ (depending on $\epsilon$) such that $$ |E(\delta x) - \ell| \leqslant \epsilon $$ for all $0<|\delta x| \leqslant \Delta$, then $E(\delta x)$ has limit $\ell$ as $\delta x$ tends to zero. The strict inequalities are important. The notion is that $E(\delta x)$ gets closer and closer to $\ell$ as $\delta x$ gets smaller (but not zero). But beware. The limit does not always exist.

All this is usually written as $$\lim_{\delta x \to 0} E(\delta x) =\ell$$ or $E(\delta x) \to \ell$ as $\delta x \to 0$.

Mathematically there then follows a process of proving all kinds of different properties of limits, such as the sum or product of two limits is the limit of the sum or product and so on.

In you example $f(x) = x^2$, so you would argue: take any $\epsilon > 0$. Then for any $\delta x$, $$ \frac{(x+\delta x)^2 - x^2}{\delta x} = 2x + \delta x $$ which leads to $$ \left|\frac{f(x+\delta x) - f(x)}{\delta x} - 2x\right| = |\delta x|.$$ Now, if you choose $\Delta = \epsilon$, for every $\delta x$ with $|\delta x| < \Delta$ you have the inequality you want and you can say $$ f'(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x)-f(x)}{\delta x} = 2x.$$ With practice and using the many of properties of limits this gets easier.

$f'(x)=2x+dx$ is technically invalid. You cannot add functions and differentials. A correct expression is $df=2x\,dx$, but that does not help.

Using small increments, you can indeed write $$0<\Delta x\ll x\implies f'(x)\approx\frac{(x+\Delta x)^2-x^2}{\Delta x}=2x+\Delta x.$$

Then,

$$f'(x)=\lim_{\Delta x\to0}(2x+\Delta x)=2x.$$

Too long to be a comment, but probably not satisfactory as an answer.

$dx$ is notation and it stands for 'differential with respect to $x$'. You would never write $f'(x) = 2x + dx$, because on the LHS, there's a derivative and on the right .. I don't even know what that is tbh. If it was something like $(2x+a)dx$ then it would be a differential. The symbol $dx$ does not appear by itself. Either way, the equality $f'(x) = 2x+dx$ is suspect.

On another angle, by definition $$ \lim _{h\to 0} \frac{f(x+h)-f(x)}{h} =: f'(x) $$ if it exists. So, we can approximate $f(x+h)$ linearly, writing $f(x+\Delta x) \approx f'(x)\Delta x + f(x)$, where $\Delta x$ is 'super super small'. But $\Delta x$ and $dx$ are completely different beast. $\Delta x$ is some fixed number, whereas $dx$ is not a number at all.