Can't figure out a step in the proof of $\cosh^{-1}x=\ln(x+\sqrt{x^2-1}), \forall x\ge1$? I can't figure out the part start from $*$ in the proof from my book:
Set $$y=\cosh^{-1}x,\,x\ge1$$
and note that $$\cosh y=x\quad and\quad y\ge0.$$
... .
$(*)$Since $y$ is nonnegative,
$$e^y=x\pm\sqrt{x^2-1}$$
cannot be less than $1$.
This renders the negative sign impossible.
... .
 A: Note that we have $$\cosh y=\frac{e^y+e^{-y}}{2}=x$$
So we have by multiplying $2e^{y}$ on each side, $$e^{2y}-2xe^{y}+1=0$$
 Setting $e^{y}=t$ and applying the quadratic formula, we have that $$e^{y}=x \pm \sqrt{x^2-1}$$
Note that $\cosh^{-1} x$ is a function that goes from the real numbers which are greater than $1$ to the non-negative reals. So we have that $y \ge 0$. Since $e^{y}$ is an increasing function, $$e^{y}\ge e^{0}=1$$
However,$$x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}} \le \frac{1}{1+\sqrt{1^2-1}}=1$$
So $e^{y} \neq x-\sqrt{x^2-1}$ given that $x \neq 1$. In the case when $x=1$, we have that $x-\sqrt{x^2-1}=x+\sqrt{x^2-1}$. 
Thus, $e^{y}=x+\sqrt{x^2-1}$. 
A: By the way, an alternative proof which avoids faffing around with exponentials: the LHS and RHS are equal at $x=1$, while $\dfrac{d}{dx} \cosh^{-1}(x)$ may be found as follows: $$1 = \dfrac{d}{dx} \cosh \cosh^{-1} x = \dfrac{d}{dx} \cosh^{-1}(x) \sinh \cosh^{-1}(x)$$
so $$\dfrac{d}{dx} \cosh^{-1}(x) = \frac{1}{\sinh \cosh^{-1}(x)}$$
Now $\cosh^2(x) - \sinh^2(x) = 1$, so $x^2 - \sinh(\cosh^{-1} x)^2 = 1$,
whereupon $$\sinh(\cosh^{-1}(x)) = \pm \sqrt{x^2-1}$$
so $$\dfrac{d}{dx} \cosh^{-1} x = \pm (x^2-1)^{\frac{1}{2}}$$
Then note that $\cosh^{-1}(x)$ is increasing on $x>1$, so we must pick the positive sign on $x > 1$. So we have differentiated the left-hand side.
On the other hand, $$\dfrac{d}{dx} \log\left(x+\sqrt{x^2-1}\right)$$ can be easily shown to be the same expression by repeated use of the chain rule.
Since the LHS and RHS are equal at the point $x=0$, and have equal derivatives, they must be equal on all points of their mutual domains.
