Com es diferencia de f (x) = sqrt (ln (1 / sqrt (xe ^ x)) utilitzant la regla de la cadena?

Resposta:

Simplement reguli la cadena una vegada i una altra.

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

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

Bé, serà difícil:

#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. Aquests exercicis han de ser il·legals.