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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 sqrt ((x + 1) / (2x-1))?
- (3 (x + 1)) / (2x2) -1 (2x1) / (2x-1)) f (x) = u ^ nf '(x) = n xx du) / dx xxu ^ (n-1) I dette tilfælde: sqrt ((x + 1) / (2x-1)) = ((x + 1) / (2x-1)) ^ n = 1/2, u = (x + 1) / (2x-1) d / dx = 1/2 xx (1xx (2x-1) - 2xx (x + 1)) / (2x-1) ^ 2 xx (x x 1) / (2x-1)) ^ (1 / 2-1) = 1 / 2xx (-3) / (2x-1) 2xx ((x + 1) / (2x- 1)) ^ (1 / 2-1) = - (3 (x + 1)) / (2 (2x-1) ^ 2 ((x + 1) / (2x-1)) ^
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