Question

Write \(\cosh \left( {4\ln x} \right)\) as a rational function of x.

Short answer

The rational function of the expression \(\cosh \left( {4\ln x} \right)\) is \(\frac{{{x^8} + 1}}{{2{x^4}}}\).

Step-by-step solution

Definition of Hyperbolic function

The formulas for the hyperbolic function as shown below:

\(\begin{aligned}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm csch}\nolimits} x = \frac{1}{{\sinh x}}\\\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm sech}\nolimits} x = \frac{1}{{\cosh x}}\\\tanh x = \frac{{\sinh x}}{{\cosh x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\coth x = \frac{{\cosh x}}{{\sinh x}}\end{aligned}\)

Write \(\cosh \left( {4\ln x} \right)\) as a rational function of x

Write the expression as a rational function of \(x\) as shown below:

\(\begin{aligned}\cosh \left( {4\ln x} \right) &= \cosh \left( {\ln {x^4}} \right)\\ &= \frac{{{e^{\ln {x^4}}} + {e^{ - \ln {x^4}}}}}{2}\\ &= \frac{{{x^4} + {x^{ - 4}}}}{2}\\ &= \frac{{{x^4} + \frac{1}{{{x^4}}}}}{2}\end{aligned}\)

Solve further,

\(\begin{aligned}\cosh \left( {4\ln x} \right) &= \frac{{\frac{{{x^8} + 1}}{{{x^4}}}}}{2}\\ &= \frac{{{x^8} + 1}}{{2{x^4}}}\end{aligned}\)

Thus, the rational function of the expression \(\cosh \left( {4\ln x} \right)\) is \(\frac{{{x^8} + 1}}{{2{x^4}}}\).