First derivative equation with $\cosh^{-1}$ set equal to zero solving This is the equation I'm left with when I took the derivative of a function. I would like to optimise so I'm trying to find a min/max by setting equal to zero. I have been having trouble solving. 
$$\frac{2\text{arcosh}(y)}{\sqrt{y^2-1}}+2y-4=0$$

 A: You want to minimize
$$
f(x)=d^2=x^2+(\cosh x-2)^2.
$$
Differentiating, we find that
$$
f'(x)=2x+2\cosh x\sinh x-4\sinh x,
$$
and that
$$
f'(0)=0.
$$
Now, 
$$
f''(x)=8\cosh x\sinh^2(x/2)\geq 0
$$
and $f''(x)=0$ if and only if $x=0$. Thus, you have a strictly concave function. Can you conclude from here?
Update
If you are not too familiar with convexity, you can argue like this:
$f'(0)=0$, and $f''(x)>0$ if $x>0$. This means that $f'$ is increasing for positive $x$. In particular $f'(x)>f'(0)=0$ if $x>0$. But if $f'(x)>0$ for all $x>0$ it means, in turn, that $f$ is increasing for positive $x$. In particular, $f(x)>f(0)=1$ if $x>0$.
On the other hand $f$ is even ($f(-x)=f(x)$), so $f(x)>f(0)$ also if $x<0$. We conclude that $f$ has a global minimum at $x=0$. 
A: $y=\cosh(x)$ is a convex function and the curvature of its graph at $(x,\cosh(x))$ equals $\cosh(x)$.
This is enough to conclude that the shortest path for Bob for reaching the catenary is to go towards its vertex: see catenary evolute. As an alternative, you may notice that the squared distance of Bob from $(x,\cosh(x))$ is given by
$$ x^2+(2-\cosh(x))^2 = 1+\sum_{m\geq 2}\frac{4^m-8}{(2m)!}x^{2m} $$
that is an increasing function of $x^2$, since $\frac{4^m-8}{(2m)!}>0$ for any $m\geq 2$.
