Shortest distance of line $y=x$ to the curve $y=\cosh x$ I used the line point distance formula and tried to minimize that
$$\frac{Ax+By+C}{\sqrt {A^2 + B^2 }}$$
 Putting $y=\cosh x$
$$\frac{Ax+B\cosh x+C}{\sqrt {A^2 + B^2 }} = G(x)$$
Using values of A and B from $y=x$
$$\frac{x-\cosh x}{\sqrt {2 }} = G(x)$$
and I minimized $G(x)$ using derivatives
$$\frac{1-\sinh x}{\sqrt {2 }} = G'(x)\implies \sinh x=1$$
but then the $x$ value for that doesn't seem to be the $x$ value which minimizes the distance.
 A: Let $y=x+n$ is a tangent line to the $y=\cosh x$, which is a parallel to $y=x$.
Thus, $$\left(\frac{e^x+e^{-x}}{2}\right)'=1.$$
Can you end it now?
I got $1-\frac{1}{\sqrt2}\ln(1+\sqrt2).$
A: You are correct, the shortest distance between the line $y=x$ and the curve $y=\cosh(x)$ is attained when $\sinh(x)=1$, that is at $$x_0=\text{arcsinh}(1)=\ln(1+\sqrt{2}).$$
which yields that the desired shortest distance is
$$d=\frac{|x_0-\cosh(x_0)|}{\sqrt {2 }}=\frac{|\ln(1+\sqrt{2})-\sqrt{2}|}{\sqrt {2 }}=1-\frac{\ln(1+\sqrt2)}{\sqrt2}\approx 0.37677.$$
A: Try to think geometrically: since $\cosh x$ is a convex function, its graph lies above any tangent line, and there are tangent lines with any slope since $\frac{d}{dx}(\cosh x)=\sinh x$ is bijective. Consider now the tangent line with unit slope, going through $(\text{arcsinh} 1,\cosh\text{arcsinh} 1)=(\log(1+\sqrt{2}),\sqrt{2})$. Its distance from the line $y=x$ is $1-\frac{1}{\sqrt{2}}\log(1+\sqrt{2})$, and this is also the answer to your problem. Can you figure out why?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Let $\ds{\mathcal{D}^{2}\pars{x,\alpha} \equiv \pars{x - \alpha}^{2} + \bracks{x - \cosh\pars{\alpha}}^{\, 2}}$
\begin{align}
\left\{\begin{array}{rcrcl}
\ds{2\pars{x - \alpha}} & \ds{+} & \ds{2\bracks{x - \cosh\pars{\alpha}}} &
\ds{=} & \ds{0}
\\
\ds{2\pars{x - \alpha}\pars{-1}} & \ds{+} &
\ds{2\bracks{x - \cosh\pars{\alpha}}\bracks{-\sinh\pars{\alpha}}} &
\ds{=} & \ds{0}
\end{array}\right.
\end{align}
In "adding" both equations, I found
$\ds{\braces{\sinh\pars{\alpha} = 1,\cosh\pars{\alpha} = \root{2}} \implies
\alpha = \operatorname{arcsinh}\pars{1} = \ln\pars{1 + \root{2}}}$. In addition, $\ds{x = {\alpha + \cosh\pars{\alpha} \over 2} =
{\ln\pars{1 + \root{2}} + \root{2} \over 2}}$.
Then,
\begin{align}
&\bbox[15px,#ffd,border:1px solid navy]{\mathcal{D}^{2}\pars{x,\alpha}}
\\[5mm] = &\
\bracks{{\ln\pars{1 + \root{2}} + \root{2} \over 2} -
\ln\pars{1 + \root{2}}}^{2}
\\ + &
\,\,\bracks{{\ln\pars{1 + \root{2}} + \root{2} \over 2} - \root{2}}^{2}
\\[5mm] = &\
{\bracks{\root{2} - \ln\pars{1 + \root{2}}}^{2} \over 2}
\\[5mm] \implies &\
\bbox[15px,#ffd,border:1px solid navy]{\mathcal{D}\pars{x,\alpha} =
1 - {\root{2} \over 2}\,\ln\pars{1 + \root{2}}}\ \approx\ 0.3768
\end{align}
