Derivation of proper time with respect to time I’m confused on how the author got this answer.
He started with this:
$$\frac{d(\text{proper time})}{dt}=\frac{\sqrt{dt^2-dx^2}}{dt}
=\sqrt{1-v^2}$$
I don’t get how this is solved, specifically where the $1$ came from in $1-v^2$.
 A: The author is doing typical "physics mathematics", so it may be considered to be a little sloppy to a pure mathematician. ;)
The $1$ comes from $\frac{dt}{dt} = 1$. Perhaps it will make more sense if we do the derivation in a more formal way.
It's common to denote proper time using, $\tau$, the Greek letter tau. Note that the author is using natural units where $c$, the speed of light, equals $1$.
Let
$$(\Delta\tau)^2 = (\Delta t)^2 - (\Delta x)^2$$
Dividing through by $(\Delta t)^2$,
$$\left(\frac{\Delta\tau}{\Delta t}\right)^2 = \left(\frac{\Delta t}{\Delta t}\right)^2 - \left(\frac{\Delta x}{\Delta t}\right)^2$$
or
$$\left(\frac{\Delta\tau}{\Delta t}\right)^2 = 1 - \left(\frac{\Delta x}{\Delta t}\right)^2$$
Taking limits as $\Delta t \to 0$,
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - \left(\frac{dx}{dt}\right)^2$$
But $\frac{dx}{dt}=v$, so
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - v^2$$
Taking square roots,
$$\frac{d\tau}{dt} = \sqrt{1 - v^2}$$
which is the desired result.

Note that
$$\gamma = \frac{1}{\sqrt{1 - v^2}}$$
is the Lorentz factor.
