Wie unterscheidet man f (x) = sqrt (cote ^ (4x) anhand der Kettenregel.)

Antworten:

#f '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (Kinderbett (e ^ (4x))) ^ (- 1/2)) / 2 #

#color (weiß) (f '(x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (Kinderbett (e ^ (4x)) #

Erläuterung:

#f (x) = sqrt (Kinderbett (e ^ (4x))) #

#color (weiß) (f (x)) = sqrt (g (x)) #

#f '(x) = 1/2 * (g (x)) ^ (- 1/2) * g' (x) #

#Farbe (weiß) (f '(x)) = (g' (x) (g (x)) ^ (- 1/2)) / 2 #

#g (x) = Kinderbett (e ^ (4x)) #

#color (weiß) (g (x)) = Kinderbett (h (x)) #

#g '(x) = - h' (x) csc ^ 2 (h (x)) #

#h (x) = e ^ (4x) #

#color (weiß) (h (x)) = e ^ (j (x)) #

#h '(x) = j' (x) e ^ (j (x)) #

#j (x) = 4x #

#j '(x) = 4 #

#h '(x) = 4e ^ (4x) #

#g '(x) = - 4e ^ (4x) csc ^ 2 (e ^ (4x)) #

#f '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (Kinderbett (e ^ (4x))) ^ (- 1/2)) / 2 #

#color (weiß) (f '(x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (Kinderbett (e ^ (4x)) #

Wie unterscheidet man f (x) = sqrt (ln (x ^ 2 + 3) anhand der Kettenregel.?

F '(x) = (x (ln (x ^ 2 + 3)) ^ (- 1/2)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3))) Wir sind gegeben: y = (ln (x ^ 2 + 3) ) ^ (1/2) y '= 1/2 * (In (x ^ 2 + 3)) ^ (1 / 2-1) * d / dx [In (x ^ 2 + 3)] y' = ( ln (x ^ 2 + 3)) ^ (- 1/2) / 2 · d / dx [ln (x ^ 2 + 3)] d / dx [ln (x ^ 2 + 3)] = (d / dx [x ^ 2 + 3]) / (x ^ 2 + 3) d / dx [x ^ 2 + 3] = 2x y '= (In (x ^ 2 + 3)) ^ (- 1/2) / 2 * (2x) / (x ^ 2 + 3) = (x (ln (x ^ 2 + 3)) ^ (- 1/2)) / (x ^ 2 + 3) = x / ((x ^ 2 +) 3) (In (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (In (x ^ 2 + 3)))

Wie unterscheidet man f (x) = sqrt (e ^ cot (x)) anhand der Kettenregel?

F '(x) == - (sqrt (e ^ cot (x)). csc ^ 2 (x)) / 2 f (x) = sqrt (e ^ cot (x)) Um die Ableitung von f (x ), müssen wir eine Kettenregel verwenden. Farbe (rot) "Kettenregel: f (g (x)) '= f' (g (x)). g '(x)" Sei u (x) = cot (x) => u' (x) = -csc ^ 2 (x) und g (x) = e ^ (x) => g '(x) = e ^ (x). g' (u (x)) = e ^ cot (x) f (x ) = sqrt (x) => f '(x) = 1 / (2sqrt (x)) => f' (g (u (x))) = 1 / (2sqrt (e ^ cot (x)) d / dx (f (g (u (x))) = f '(g (u (x))). g' (u (x)). u '(x) = 1 / (sqrt (e ^ cot (x ))) e ^ cot (x) - cos ^ 2 (x) = (- e ^ cot (x) csc ^ 2x) / sqrt (e

Wie unterscheidet man f (x) = sqrt (ln (1 / sqrt (xe ^ x)) anhand der Kettenregel?

Kettenregel immer und immer wieder. 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))) Okay, das wird hart sein: 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