Odgovor:
Pojasnilo:
Pravilo verige:
Najprej ločite zunanjo funkcijo, pri čemer pustite notranjost samo, nato pa jo pomnožite z izpeljano notranjo funkcijo.
#y = tan sqrt (3x-1) #
# dy / dx = sec ^ 2 sqrt (3x-1) * d / dx sqrt (3x-1) #
# = sek ^ 2 sqrt (3x-1) * d / dx (3x-1) ^ (1/2) #
# = sek ^ 2 sqrt (3x-1) * 1/2 (3x-1) ^ (- 1/2) * d / dx (3x-1) #
# = sek ^ 2 sqrt (3x-1) * 1 / (2 sqrt (3x-1)) * 3 #
# = (3 sec ^ 2 sqrt (3x-1)) / (2 sqrt (3x-1)) #
Kako najdete derivat f (x) = [(2x-5) ^ 5] / [(x ^ 2 + 2) ^ 2] z uporabo verižnega pravila?
![Kako najdete derivat f (x) = [(2x-5) ^ 5] / [(x ^ 2 + 2) ^ 2] z uporabo verižnega pravila?](https://img.go-homework.com/precalculus/how-do-you-find-the-end-behavior-of-a-quadratic-function-3.jpg "Kako najdete derivat f (x) = [(2x-5) ^ 5] / [(x ^ 2 + 2) ^ 2] z uporabo verižnega pravila?")
= (10 (2x-5) ^ 4 * (x ^ 2 + 2) ^ 2 - (2x-5) ^ 5 * 4x (x ^ 2 + 2)) / (x ^ 2 + 2) ^ 4 f ' (x) = (f '(x) * g (x) - f (x) * g' (x)) / (g (x)) ^ 2 f '(x) = (((5 (2x-5) ) ^ 4 * 2) (x ^ 2 + 2) ^ 2) - (2x-5) ^ 5 * (2 (x ^ 2 + 2) * 2x)) / ((x ^ 2 + 2) ^ 2) ^ 2 = (10 (2x-5) ^ 4 * (x ^ 2 + 2) ^ 2 - (2x-5) ^ 5 * 4x (x ^ 2 + 2)) / (x ^ 2 + 2) ^ 4 Lahko zmanjšate več, vendar je dolgočasno rešiti to enačbo, uporabite algebraično metodo.
Kako ločite f (x) = tan (e ^ ((lnx-2) ^ 2)) z uporabo verižnega pravila.
((2sec ^ 2 (e ^ ((ln (x) -2) ^ 2)) e ^ ((ln (x) -2) ^ 2) (lnx-2)) / x) d / dx (tan ( e ^ ((ln (x) -2) ^ 2))) = sec ^ 2 (e ^ ((ln (x) -2) ^ 2)) * d / dx ((e ^ ((ln (x) -2) ^ 2)) = sek ^ 2 (e ^ ((ln (x) -2) ^ 2)) e ^ (((ln (x) -2)) ^ 2) * d / dx (ln ( x) -2) ^ 2 = sek ^ 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 )
Kako ločite f (x) = sqrt (ln (1 / sqrt (xe ^ x)) z uporabo verižnega pravila.?
Samo z verigo vedno znova. 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))) Ok, to bo težko: 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)) '= = sqr