대답:
그냥 규칙을 반복해서 반복하십시오.
(xe ^ x) ^ 3)) #f '(xe ^ x) / eq (xe ^ x)
설명:
#f (x) = sqrt (ln (1 / sqrt (xe ^ x))) #
좋아, 이건 힘들거야.
= (sqrt (ln (1 / sqrt (xe ^ x)))) '= #
(1 / sqrt (xe ^ x))) '= # 1 / (2sqrt (ln (1 / sqrt (xe ^ x)
(1 / sqrt (xe ^ x)) '= # 1 / (1 / sqrt (xe ^ x)))
* sqrt (xe ^ x) (1 / sqrt (xe ^ x)) '= # 1 / (2sqrt (ln (1 / sqrt
# = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))) (1 / sqrt (xe ^ x)) '= #
(xe ^ x) ^ - (1/2)) '= # # sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt
(xe ^ x) ^ - (3/2)) (xe ^ x) = (xe ^ x) / (2sqrt (ln (1 / sqrt '= #
(xe ^ x) ^ - (3/2)) (xe ^ x) '= # # sqrt (xe ^ x) / (4sqrt (ln (1 / sqrt
(xe ^ x) '= # # sqrt (xe ^ x)
xe ^ x) '= # # sqrt (xe ^ x) / (4sqrt (ln (1 / sqrt (xe ^ x)
(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 = #
(e ^ x + xe ^ x) = # 1 / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)
(xe ^ x) ^ 3)) # = e ^ x (1 + x) / 4sqrt (xe ^ x) / (ln (1 / sqrt
추신 이러한 운동은 불법입니다.
