Antworten:
Erläuterung:
In diesem Fall:
Wie unterscheidet man f (x) = sqrt (cote ^ (4x) anhand der Kettenregel.)
F '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (Kinderbett (e ^ (4x))) ^ (- 1/2)) / 2 Farbe (weiß) (f' (x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (Kinderbett (e ^ (4x)) f (x) = sqrt (Kinderbett (e ^ (4x))) Farbe (weiß) (f (x)) = Quadrat (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) = Bett (e ^ (4x)) Farbe (weiß) (g) (x)) = cot (h (x)) g '(x) = - h' (x) csc ^ 2 (h (x)) h (x) = e ^ (4x) Farbe (weiß) (h ( x)) = e ^ (j (x)) h '(x) = j' (x) e ^ (j (x)) j (x) = 4x j '(x) = 4 h' (x) = 4e ^ (4x) g
Wie unterscheidet man sqrt (cos (x ^ 2 + 2)) + sqrt (cos ^ 2x + 2)?
(dy) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy ) / (dx) = 1 / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) sen (x ^ 2 + 2) * 2x + 2sen (x + 2) (dy ) / (dx) = (2xsen (x ^ 2 + 2) + 2sen (x + 2)) / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy) / (dx) = (cancel2 (xsen (x ^ 2 + 2) + sen (x + 2))) / (cancel2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dy) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2)))
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