Paano mo iibahin ang sqrt (cos (x ^ 2 + 2)) + sqrt (cos ^ 2x + 2)?

Sagot:

(x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) # # (dy) / (dx)

Paliwanag:

(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 ^

(x + 2))) / (cancel2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) #

(x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) # # (dy) / (dx)

Paano mo iibahin ang f (x) = sqrt (cote ^ (4x) gamit ang tuntunin ng kadena.

(x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (cot (e ^ (4x))) ^ (- 1/2)) / (x) = sqrt (cot (e ^ (4x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (cot (e ^ (4x) (x) = sqrt (g (x)) f '(x) = 1/2 * (g (x)) ^ (- 1/2) * g' (x) 2 g (x) = cot (e ^ (4x)) kulay (puti) (g) (x)) = cot (h (x)) g '(x) = - h' (x) csc ^ 2 (h (x)) h (x) = e ^ (4x) (x) = h (x) = j '(x) e ^ (j (x) = j (x) = 4x j' (x) = 4 h ' 4e ^ (4x) g '(x) = - 4e ^ (4x) csc ^ 2 (e ^ (4x)) f' (x) = (- 4e ^ (4x) csc ^ 2 (e ^ (4x)) (cot (e ^ (4x))) ^ (- 1/2)) / 2 kulay (puti) (f '(x)) = - (2e ^ (4x) csc ^ 2 (e ^ (4x))) / sqrt (cot (e ^ (4x))

Paano mo gawing simple (1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) div sqrt (a + 1) / ( (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1)), isang> 1?

Malaking format ng math ...> kulay (asul) (((1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1) (a / 1) sqrt (a + 1) - (a + 1) sqrt (a-1))) = kulay (pula) (((1 / sqrt (a- (A + 1)) / (sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a-1) Cdot sqrt (a-1) cdot sqrt (a + 1) -sqrt (a + 1) cdot sqrt (a + 1) sqrt (a-1) asul) (/ (sqrt (a-1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (A-1) (sqrt (a-1) -sqrt (a + 1))) = kulay (pula) (A / 1) sqrt (a-1) sqrt (a + 1) (sqrt (a-1) -sqrt (a + 1))) / sqrt (a + 1) = kulay (asul) ((1 / sqrt (a + 1)) xx ((sqrt (a + 1) cdot sqrt (a-1)) / (sqr

Paano mo iibahin ang f (x) = sqrt (ln (1 / sqrt (xe ^ x)) gamit ang tuntunin ng kadena.

Iisa lamang ang panuntunan. f '(x) = e ^ x (1 + x) / 4sqrt ((xe ^ x) / (ln (1 / sqrt (xe ^ x)) (xe ^ x) ^ 3) (ln (1 / sqrt (xe ^ x))) Okay, ito ay magiging mahirap: f '(x) = (sqrt (ln (1 / sqrt (xe ^ x) (ln (1 / sqrt (xe ^ x))))) * (ln (1 / sqrt (xe ^ x))) '= = 1 / (2sqrt (ln (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) (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))) ((xe ^ x) ^ - (1/2)) '= = sqrt (xe ^ x) / (2sqrt (ln (1 / sqrt (xe ^ x)))))) (- 1/2) ((xe ^ x) ^ - (3/2)) (xe ^ x) '= = sqrt (xe ^ x) / (4