# dy/dx=sinh(x) A tangent line through the origin has equation y=mx. If it meets the graph at x=a, then ma=cosh(a) and m=sinh(a). Therefore, asinh(a)=cosh(a). Use Newton's Method to solve for a

$dy/dx=\mathrm{sinh}\left(x\right)$ A tangent line through the origin has equation $y=mx$. If it meets the graph at $x=a$, then $ma=\mathrm{cosh}\left(a\right)$ and $m=\mathrm{sinh}\left(a\right)$. Therefore, $a\mathrm{sinh}\left(a\right)=\mathrm{cosh}\left(a\right)$.
Use Newton's Method to solve for $a$
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You are looking for the zero of function
$f\left(a\right)=a\mathrm{sinh}\left(a\right)-\mathrm{cosh}\left(a\right)$
for which
${f}^{\prime }\left(a\right)=a\mathrm{cosh}\left(a\right)\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{f}^{″}\left(a\right)=a\mathrm{sinh}\left(a\right)+\mathrm{cosh}\left(a\right)$
Since the function is even, you have two symmetric solutions.
What you can notice is that$f\left(0\right)=-1$, ${f}^{\prime }\left(0\right)=0$ and ${f}^{″}\left(0\right)=1$. So, to generate a guess, perform a Taylor expansion around $a=0$; this would give
$f\left(a\right)=-1+\frac{{a}^{2}}{2}+\frac{{a}^{4}}{8}+O\left({a}^{6}\right)$
which is a quadratic in ${a}^{2}$. So, an estimate is
${a}_{0}=\sqrt{2\left(\sqrt{3}-1\right)}$
Now, use Newton method which generates as itegrates
${a}_{n+1}={a}_{n}-\frac{f\left({a}_{n}\right)}{{f}^{\prime }\left({a}_{n}\right)}={a}_{n}+\frac{1}{{a}_{n}}-\mathrm{tanh}\left({a}_{n}\right)$
which should converge very fast.