How to solve transcendental hyperbolic equation

Question

How can I solve the functional relation $$ e^{-af'(x)}\cosh( f(x) ) = bx $$ for $f(x)$? It would suffice to solve for $x>0$, $a>0$ and $b>0$.

Answer 1

for $a$: $-{\frac {1}{f \left( x \right) }\ln \left( {\frac {bx}{\cosh \left( { \frac {\rm d}{{\rm d}x}}f \left( x \right) \right) }} \right) } $ for $b$: ${\frac {{{\rm e}^{-af \left( x \right) }}\cosh \left( {\frac {\rm d}{ {\rm d}x}}f \left( x \right) \right) }{x}} $ for $x$ no chance