Jawapan:
Penjelasan:
#f '(x) = sec (e ^ x-3x) tan (e ^ x-3x) terbitan (e ^ x-3x)
Bagaimana anda membezakan f (x) = sqrt (cote ^ (4x) menggunakan peraturan rantai.?
(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (cot (e ^ (4x))) ^ (- 1/2) (x) = sqrt (cot (e ^ (4x)) = - (2e ^ (4x) csc ^ (x) = f (x)) = sqrt (g (x)) f '(x) = 1/2 * (g (x)) ^ (- 1/2) (g '(x)) = (g' (x) (g (x)) ^ (- 1/2)) / 2 g (x) = cot (e ^ (4x) (x) = cot (h (x)) g '(x) = - h' (x) csc ^ 2 (h (x) (x) = h (x) = j '(x) e ^ (j (x) = j (x) = 4x j' (x) = 4 h ' (X) = - 4e ^ (4x) csc ^ 2 (e ^ (4x)) f '(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x) (cot (e ^ (4x))) ^ (- 1/2)) / 2 warna (putih) (f '(x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (cot (e ^ (4x))
Bagaimana anda membezakan f (x) = sqrt (ln (x ^ 2 + 3) menggunakan peraturan rantai.?
(x ^ 2 + 3) = x / ((x ^ 2 + 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (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 [ (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)) / 3) (ln (x ^ 2 + 3)) ^ (1/2)) = x / ((x ^ 2 + 3) sqrt (ln (x ^ 2 + 3)))
Bagaimana anda membezakan f (x) = (x ^ 3-2x + 3) ^ (3/2) menggunakan peraturan rantai?
(3x ^ 2 - 2) Peraturan rantai: d / dx f (g (x)) = f '(g (x)) * g' (x) Peraturan kuasa: d / dx x ^ n = n * x ^ (n-1) Menerapkan peraturan ini: 1 Fungsi dalaman, g (x) ialah x ^ 3-2x + (x) adalah g (x) ^ (3/2) 2 Ambil derivatif fungsi luar dengan menggunakan kuasa kuasa d / dx (g (x)) ^ (3/2) = 3/2 * g (x) ^ (3/2 - 2/2) = 3/2 * g (x) ^ (1/2) = 3/2 * sqrt (g (x)) f '(g (x) (x ^ 3 - 2x + 3) 3 Ambil derivatif fungsi dalaman d / dx g (x) = 3x ^ 2 -2 g '(x) = 3x ^ 2 -2 4 Multiply f' ) dengan penyelesaian g '(x) (3/2 * sqrt (x ^ 3 - 2x + 3)) * (3x ^ 2 - 2): 3/2 * (sqrt (x ^ 3 - 2x + 3) * (3x ^ 2 - 2)