Question on $\frac{dy}{dx}$ behaving like a fraction So, $\frac{dy}{dx}$ is defined as a limit. However, because of the
chain rule: $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
And inverse function theorem: $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$
It seems that $\frac{dy}{dx}$ has "fraction like" behavior.
Could somebody explain to me what justifies the second equality?
length ds = $\sqrt{(dx)^2+(dy)^2}=\sqrt{{1+(\frac{dy}{dx})^2}}dx$
Page 573 equation (1) of Gilbert Strang's calculus text
 A: The expression $\sqrt{dx^2+dy^2}$ is shorthand - you can’t typically combine differentials like this and what would it even mean to integrate that? What it means is that if $x$ and $y$ are variables of a curve that are parametrized by some $t$, then to integrate $\sqrt{\frac{dx}{dt}^2+ \frac{dy}{dt}^2}dt$. If you factor out the $\frac{dx}{dt}$ term, you can apply the chain rule to the inside using $\frac{dy}{dt}/\frac{dx}{dt}= \frac{dy}{dx}$. On the outside, you can do u-substitution on your integral to get $\frac{dx}{dt}dt=dx$, giving you your alternative expression.
A: I had a quick look at Gilbert's text and here is what I understood:
You are given a curve $\gamma=(x,y):I\to\mathbb R^2$, where $I\subset\mathbb R$ is some open interval.
Suppose for example that the curve is smooth and that $\gamma'(t)\neq 0$ for all $t\in I$.
Then as derived by me here, the length of $\gamma$ is given by $$\int_I \|\gamma'(t)\| \,\mathrm dt.$$ If you use the strange-looking notation $\gamma' = (\mathrm dx, \mathrm dy)$, then you obtain the "$\mathrm ds = \sqrt{\mathrm dx^2+\mathrm dy^2}$" part. (I will maybe give a justification for this notation down below in the future, but currently that part is incomplete.)
Furthermore, an application of the implicit function Theorem [2; Satz 169.1], details left as exercise to you, gives that there exists (after relabeling the axes $x$ and $y$ if necessary) for each $t\in I$ a neighborhood $J\subset I$ and a smooth function $f_J:J\to\mathbb R$ such that $\gamma(t) = (x(t),f_J(x(t)))$ for all $t\in J$.
Then the length of $\gamma\vert_J$ is $$\int_J \|\gamma'(t)\| \,\mathrm dt = \int_J \|(x(t),f_J(x(t)))'\| \,\mathrm dt=\int_J\lvert x'(t)\rvert\sqrt{1+f_J'(x(t))^2}\,\mathrm dt.$$ Now substitute $u=x=x(t)$ to obtain, in your notation, "$\mathrm ds = \sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$". (Here, $\frac{\mathrm dy}{\mathrm dx}$ is short for $f_J'$.) (And indeed you have to be wary of the sign of $x'$.)

This part is not finished yet.
Then every point $t\in I$ has a neighborhood $J\subset I$ such that the image of $\gamma\vert_J$ is an embedded $1$-dimensional smooth submanifold of $\mathbb R^2$. This follows from the Local Parametrization Theorem [1; Theorem 2.5]. Indeed,  Such a $f_J$ provides the local parametrizations for [1; Theorem 2.5].

Literature
[2] Harro Heuser, Lehrbuch der Analysis. Teil 2. 11. Auflage. 2000.
A: $$ \sqrt{a^2+b^2} = \sqrt{a^2\left(1+\frac{b^2}{a^2}\right)}=a\sqrt{1+\frac{b^2}{a^2}} \quad \text{with}\ \ a=dx\ \ \text{and}\ \ b = dy.$$
You would need to justify why $\ dx\geq 0\ $ for the second equality above to hold, but that's not the focus of your question...
$$$$
Further to questions in the comments, let's say you want to calculate the (arc) length of $\ y=x^2\ $ between $\ x=0\ $ and $\ x=1.$ Then see here for the formula and how it is derived.
