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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.

... .

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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, $e^{y}=x+\sqrt{x^2-1}$.

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    @N1ng Do you understand? – 2017-02-19
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    I'm thinking the $\le$-sign following the $However$. – 2017-02-19
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    If I change that part following $However,$ with $x+\sqrt{x^2-1}$, then I may get the other conclusion that $e^y\ne x+\sqrt{x^2-1}$, given that $x\ne 1$? – 2017-02-19
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    @N1ng No, you cannot. You have $x-\sqrt{x^2-1} <1$ for $x >1$, which means that $e^{y} \neq x-\sqrt{x^2-1} $ for $x>1$. Also, if we chang ethe part following However, note that we have that $$x+\sqrt{x^2-1}=\frac{1}{x-\sqrt{x^2-1}} \ge 1$$ as $x-\sqrt{x^2-1}$ is a decreasing graph for $x \ge 1$ and $x+\sqrt{x^2-1}$ is an increasing graph for $x \ge 1$. – 2017-02-19
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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.

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    How a creative way to proof, thanks. I take it as $\pm(x^2-1)^{\frac{-1}{2}}$ on line-$9$. But I'm still confused that you said $\cosh^{-1}(x)$ is decreasing on $x\lt 0$, since by definition $x\ge 1$. – 2017-02-19
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    Oh, my bad. You're quite right, I was thinking of $\cosh$ :P – 2017-02-19