Kako ločite y = ln (e ^ x + sqrt (1 + e ^ (2x)))?

Odgovor:

# (dy) / (dx) = (e ^ x) / (sqrt (1 + e ^ (2x))) #

Pojasnilo:

Uporabite pravilo verige.

#u (x) = e ^ x + (1 + e ^ (2x)) ^ (1/2) in y = ln (u) #

# (dy) / (du) = 1 / u = 1 / (e ^ x + (1 + e ^ (2x)) ^ (1/2)) #

# (du) / (dx) = e ^ x + d / (dx) ((1 + e ^ (2x)) ^ (1/2)) #

Za kvadratni koren ponovno uporabite pravilo verige z

#phi = (1 + e ^ (2x)) ^ (1/2) #

#v (x) = 1 + e ^ (2x) in phi = v ^ (1/2) #

# (dv) / (dx) = 2e ^ (2x) in (dphi) / (dv) = 1 / (2sqrt (v)) #

# (dphi) / (dx) = (dphi) / (dv) (dv) / (dx) = (e ^ (2x)) / (sqrt (1 + e ^ (2x))) #

#torej (du) / (dx) = e ^ x + (e ^ (2x)) / (sqrt (1 + e ^ (2x))) #

# (dy) / (dx) = (dy) / (du) (du) / (dx) #

# = 1 / (e ^ x + (1 + e ^ (2x)) ^ (1/2)) * (e ^ x + (e ^ (2x)) / (sqrt (1 + e ^ (2x)))) #

# = e ^ x / (e ^ x + sqrt (1 + e ^ (2x))) + e ^ (2x) / (sqrt (1 + e ^ (2x)) (e ^ x + sqrt (1 + e) ^ (2x))) #

Združevanje prek zaslona LCD:

# = (e ^ xsqrt (1 + e ^ (2x)) + e ^ (2x)) / (sqrt (1 + e ^ (2x)) (e ^ x + sqrt (1 + e ^ (2x))) #

Vzemite faktor # e ^ x # iz števca:

# = (e ^ x (sqrt (1 + e ^ (2x)) + e ^ x)) / (sqrt (1 + e ^ (2x)) (e ^ x + sqrt (1 + e ^ (2x))) #

Prekličite in pridobite

# = (e ^ x) / (sqrt (1 + e ^ (2x))) #