What is #cosh(ln(x))#?
1 Answer
Nov 7, 2015
#cosh(ln(x)) = (x^2+1)/(2x)#
Explanation:
#cosh(z) = (e^z+e^-z)/2#
So:
#cosh(ln(x)) = (e^(ln(x))+e^(-ln(x)))/2 = (x+1/x)/2 = (x^2+1)/(2x)#
