Paano mo iibahin ang f (x) = sqrt (ln (1 / sqrt (xe ^ x)) gamit ang tuntunin ng kadena.

Sagot:

Iisa lamang ang panuntunan.

#f '(x) = e ^ x (1 + x) / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)

Paliwanag:

#f (x) = sqrt (ln (1 / sqrt (xe ^ x))) #

Okay, ito ay magiging mahirap:

#f '(x) = (sqrt (ln (1 / sqrt (xe ^ x))))' = #

# = 1 / (2sqrt (ln (1 / sqrt (xe ^ x)))) * (ln (1 / sqrt (xe ^ x))) '= #

# 1 = (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) '= #

(xe ^ x)))

# = sqrt (xe ^ x) / (4sqrt (ln (1 / sqrt (xe ^ x)))) 1 / sqrt (xe ^ x) ^ 3)

# = 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) '= #

(X) ^ 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. Ang mga pagsasanay na ito ay dapat na ilegal.