Answer
$\frac{{{x^8} + 1}}{{2{x^4}}}$
Work Step by Step
$$\eqalign{
& \cosh \left( {4\ln x} \right) \cr
& {\text{Use the definition of hyperbolic functions }} \cr
& \cosh t = \frac{{{e^t} + {e^{ - t}}}}{2},{\text{ Let }}t = 4\ln x \cr
& \cosh \left( {4\ln x} \right) = \frac{{{e^{4\ln x}} + {e^{ - 4\ln x}}}}{2} \cr
& {\text{Using logarithmic properties }}n\ln x = {x^{ - n}} \cr
& \cosh \left( {4\ln x} \right) = \frac{{{e^{\ln {x^4}}} + {e^{\ln {x^{ - 4}}}}}}{2} \cr
& {\text{Where }}{e^{\ln t}} = t \cr
& \cosh \left( {4\ln x} \right) = \frac{{{x^4} + \frac{1}{{{x^4}}}}}{2} \cr
& {\text{Simplifying}} \cr
& \cosh \left( {4\ln x} \right) = \frac{{\frac{{{x^8} + 1}}{{{x^4}}}}}{2} \cr
& \cosh \left( {4\ln x} \right) = \frac{{{x^8} + 1}}{{2{x^4}}} \cr} $$
