Hoe onderscheid je f (x) = sqrt (e ^ cot (x)) met behulp van de kettingregel?

Antwoord:

f '(x) == -# (Sqrt (e ^ kinderbed (x)). Csc 2 ^ (x)) / 2 #

Uitleg:

#f (x) = sqrt (e ^ kinderbed (x)) #

Om de afgeleide van f (x) te vinden, moeten we een kettingregel gebruiken.

#color (rood) "kettingregel: f (g (x)) '= f' (g (x)). g '(x)" #

Laat #u (x) = kinderbed (x) => u '(x) = - ^ 2 csc (x) #

en # g (x) = e ^ (x) => g '(x) = e ^ (x).g' (u (x)) = e ^ wieg (x) #

#f (x) = sqrt (x) => f '(x) = 1 / (2sqrt (x)) => f (g (u (x))) = 1 / (2sqrt (e ^ kinderbed (x)) #

# D / dx (f (g (u (x))) = f (g (u (x))). G '(u (x)). U' (x) #

=# 1 / (sqrt (e ^ kinderbed (x))) e ^ kinderbed (x).- cos ^ 2 (x) #

=# (- e ^ kinderbed (x) csc ^ 2x) / sqrt (e ^ kinderbed (x)) #

#color (blauw) "annuleer de e ^ wieg (x) met sqrt (e ^ wieg (x)) in de noemer" #

=-# (Sqrt (e ^ kinderbed (x)). Csc 2 ^ (x)) / 2 #

Hoe onderscheid je f (x) = sqrt (cote ^ (4x) met behulp van de kettingregel.?

F '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (ledikant (e ^ (4x))) ^ (- 1/2)) / 2 kleur (wit) (f' (x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (cot (e ^ (4x)) f (x) = sqrt (cot (e ^ (4x))) kleur (wit) (f (x)) = sqrt (g (x)) f '(x) = 1/2 * (g (x)) ^ (- 1/2) * g' (x) kleur (wit) ) (f '(x)) = (g' (x) (g (x)) ^ (- 1/2)) / 2 g (x) = ledikant (e ^ (4x)) kleur (wit) (g (x)) = ledikant (h (x)) g '(x) = - h' (x) csc ^ 2 (h (x)) h (x) = e ^ (4x) kleur (wit) (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 ^ (4

Hoe onderscheid je f (x) = sqrt (ln (x ^ 2 + 3) met behulp van de kettingregel.?

F '(x) = (x (ln (x ^ 2 + 3)) ^ (- 02/01)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3))) We krijgen: y = (ln (x ^ 2 + 3) ) ^ (1/2) y '= 1/2 * (ln (x ^ 2 + 3)) ^ (1 / 2-1) * d / dx [ln (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 '= (ln (x ^ 2 + 3)) ^ (- 1/2) / 2 * (2x) / (x ^ 2 + 3) = (x (ln (x ^ 2 + 3)) ^ (- 02/01)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3)))

Hoe onderscheid je f (x) = sqrt (ln (1 / sqrt (xe ^ x)) met behulp van de kettingregel.?

Regel maar regel opnieuw en opnieuw. 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))) Oké, dit wordt moeilijk: 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) ^