Hvordan differentierer du givet y = (secx ^ 3) sqrt (sin2x)?

Svar:

# Dy / dx = secx ^ 3 ((cos2x) / sqrt (sin2x) + 3x ^ 2tanx ^ 3sqrt (sin2x)) #

Forklaring:

Vi har # Y = uv # hvor # U # og # V # er begge funktioner af #x#.

# Dy / dx = uv '+ vu' #

# U = secx ^ 3 #

# U '= 3x ^ 2secx ^ 3tanx ^ 3 #

# V = (sin2x) ^ (1/2) #

#v '= (sin2x) ^ (- 1/2) / 2 * d / dx sin2x = (sin2x) ^ (- 1/2) / 2 * 2cos2x = (cos2x) / sqrt (sin2x) #

# Dy / dx = (secx ^ 3cos2x) / sqrt (sin2x) + 3x ^ 2secx ^ 3tanx ^ 3sqrt (sin2x) #

# Dy / dx = secx ^ 3 ((cos2x) / sqrt (sin2x) + 3x ^ 2tanx ^ 3sqrt (sin2x)) #

Hvordan differentierer du 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)) ) / (dx) = 1 / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2)) * sen (x ^ 2 + 2) * 2x + 2sen (x + 2) ) / (dx) = (2xsen (x ^ 2 + 2) + 2sen (x + 2)) / (2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) (dx) = (annuller2 (xsen (x ^ 2 + 2) + sen (x + 2))) / (annullér2sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2))) / (dx) = (xsen (x ^ 2 + 2) + sen (x + 2)) / (sqrtcos (x ^ 2 + 2) + sqrt (cos ^ 2 (x + 2)))

Hvordan forenkler du (1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) div sqrt (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1)), a> 1?

Kæmpe matematisk formatering ...> farve (blå) ((1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) ) / (sqrt (a + 1) / (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1))) = farve (rød) 1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a-1))) / +1) / (sqrt (a-1) cdot sqrt (a-1) cdot sqrt (a + 1) -sqrt (a + 1) cdot sqrt (a + 1) sqrt (a-1))) = farve blå) ((1 / sqrt (a-1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt -1)))) (sqrt (a + 1) / (sqrt (a + 1) cdot sqrt (a-1) (sqrt (a-1) -sqrt (a + 1))) = farve / (Sqrt (a-1) -sqrt (a + 1)) / (sqrt (a +

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