Hvordan differentierer du f (x) = sqrt (e ^ cot (x)) ved hjælp af kædelegemet?

Svar:

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

Forklaring:

#F (x) = sqrt (e ^ barneseng (x)) #

For at finde derivatet af f (x), skal vi bruge kæderegel.

#color (rød) "kæderegel: f (g (x)) '= f' (g (x)).g '(x)" #

Lade #u (x) = barneseng (x) => u '(x) = - csc ^ 2 (x) #

og # 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 ^ barneseng (x)) #

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

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

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

#color (blå) "annuller e ^ cot (x) med sqrt (e ^ cot (x)) i nævneren" #

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

Hvordan differentierer du f (x) = sqrt (cote ^ (4x) ved hjælp af kædelegemet.?

F '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (cot (e ^ (4x))) ^ (- 1/2)) / 2 farve (hvid) (x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (barneseng (e ^ (4x)) f (x) = sqrt farve (hvid) (f (x)) = sqrt (g (x)) f '(x) = 1/2 * (g (x)) ^ (- 1/2) * g' ) (f '(x)) = (g' (x) (g (x)) ^ (- 1/2)) / 2 g (x) = barneseng (e ^ (4x)) farve (hvid) (x)) = cot (h (x)) g '(x) = - h' (x) csc ^ 2 (h (x)) h (x) = e ^ x)) = e ^ (j (x)) h '(x) = j' (x) e ^ (j (x)) j (x) = 4x j ' 4e) (4x) g '(x) = - 4e ^ (4x) csc ^ 2 (e ^ (4x)) f' (x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (cot (e ^ (4x))) (1/2)) / 2 farve (hvi

Hvordan differentierer du f (x) = sqrt (ln (x ^ 2 + 3) ved hjælp af kædelegemet.?

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))) Vi gives: 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)) ^ (- 1/2)) / (x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3)))

Hvordan differentierer du f (x) = sqrt (ln (1 / sqrt (xe ^ x)) ved hjælp af kædelegemet.?

Bare kæde regel igen og igen. f (x) = e ^ x (1 + x) / 4sqrt (xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3)) f (x) = sqrt Ok, det bliver det svært: 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))) (1 / sqrt (xe ^ x)) = = sqrt (xe ^ x) ^ - (1/2)) = = sqrt (xe ^ x) / (2sqrt (ln) (Xe ^ x) ^ - (3/2)) (xe ^ x) '= = sqrt (xe ^ x) / (4sqrt ln (1 / sqrt (xe ^ x)))) (xe ^ x) ^ - (3/2)) (xe ^ x