Odgovor:
Samo z verigo vedno znova.
#f '(x) = e ^ x (1 + x) / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) #
Pojasnilo:
#f (x) = sqrt (ln (1 / sqrt (xe ^ x))) #
Ok, to bo težko:
#f '(x) = (sqrt (ln (1 / sqrt (xe ^ x))))') #
# = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * (ln (1 / sqrt (xe ^ x))) '= #
# = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * 1 / (1 / sqrt (xe ^ x)) (1 / sqrt (xe ^ x)) '= #
# = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * sqrt (xe ^ x) (1 / sqrt (xe ^ x)) '= #
# = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))) (1 / sqrt (xe ^ x)) '= #
# = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x))))) ((xe ^ x) ^ - (1/2)) '= #
# = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))) (- 1/2) ((xe ^ x) ^ - (3/2)) (xe ^ x) '= #
# = sqrt (xe ^ x) / (4sqrt (ln (1 / sqrt (xe ^ x)))) ((xe ^ x) ^ - (3/2)) (xe ^ x) '= #
# = sqrt (xe ^ x) / (4sqrt (ln (1 / sqrt (xe ^ x))))) 1 / sqrt ((xe ^ x) ^ 3) (xe ^ x) '= #
# = sqrt (xe ^ x) / (4sqrt (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) (xe ^ x) '= #
# = 1 / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) (xe ^ x) '= #
# = 1 / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) (x) 'e ^ x + x (e ^ x)' = #
# = 1 / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) (e ^ x + xe ^ x) = #
# = e ^ x (1 + x) / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) #
P.S. Te vaje bi morale biti nezakonite.
