If f(x) = sinh(ln x), then f '(x) =?

Question

possible solutions.....

a. (x^2 + 1)/(2x^2)

b. (x^2 - 1)/(2x^2)

c. 4x/[(x^2 + 1)^2]

d. - 4x/[(x^2 + 1)^2]

e. (2 - 2x^2) /[(x^2 + 1)^2]

f. - (2 + 2x^2) /[(x^2 + 1)^2]

g. none of these

Answer 1

Remember that sinh(x) = ( e^x - e(-x) ) / 2... So sinh(lnx) is ( e^lnx - e(-lnx) ) / 2 = x/2 - 1/(2x). This is pretty straightforward to differentiate:

f'(x) = 1/(2x^2) + 1/2 = (1+x^2)/(2x^2)
(answer a.)