I have this question and i figured an answer but I am not sure if its completely right... Someone help?
the question is
if f(x) = cosh(ln(x)), then show that f ' (x) = 1/2( (x^2 - 1) / x^2)
The way i went about it was i turned the last part into
- 1/2 ( ((x^2)/x^2) - ( 1/x^2) )
- 1/2(1- (1/x^2)
then i took the derivative of f(x)
f ' (x) = sinh(lnx) (1/x)
= (e ^ lnx - e ^ -lnx) / 2 * 1/x
= (x - x^ - 1) / 2 * 1/x
= 1/2 (1 - x^ -2)
= 1/2 ( 1 - (1 / x^2) )
Thanks for the help...