derivative of cosh
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) )
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...
Your solution is correct.
Originally Posted by Cucc1137
the answer is right, however you can use the definition of cosh before
there you donīt use de the quotient rule