odpoveď:
Len reťazec pravidlo znova a znova.
# F '(x) = e ^ x (1 + x) / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) #
vysvetlenie:
# F (x) = sqrt (ln (1 / sqrt (xe ^ x))) #
Dobre, bude to ťažké:
# 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)) #
PS: Tieto cvičenia by mali byť nezákonné.
