Miten erotat f (x) = tan (e ^ ((lnx-2) ^ 2)) ketjun sääntöä käyttäen.?

Vastaus:

# ((2sec ^ 2 (e ^ ((ln (x) -2) ^ 2)) e ^ ((ln (x) -2) ^ 2) (lnx-2)) / x) #

Selitys:

# d / dx (tan (e ^ ((ln (x) -2) ^ 2))) = sek ^ 2 (e ^ ((ln (x) -2) ^ 2)) * d / dx ((e ^ ((ln (x) -2) ^ 2)) #

=# sec ^ 2 (e ^ ((ln (x) -2) ^ 2)) e ^ (((ln (x) -2)) ^ 2) * d / dx (ln (x) -2) ^ 2 #

=# sec ^ 2 (e ^ ((ln (x) -2) ^ 2)) e ^ (((ln (x) -2)) ^ 2) 2 (lnx-2) * d / dx (lnx-2) #

=# (sek ^ 2 (e ^ ((ln (x) -2) ^ 2)) e ^ (((ln (x) -2)) ^ 2) 2 (lnx-2) * 1 / x) #

=# ((2sec ^ 2 (e ^ ((ln (x) -2) ^ 2)) e ^ ((ln (x) -2) ^ 2) (lnx-2)) / x) #