Hvordan differentierer du sqrt (cos (x ^ 2 + 2)) + sqrt (cos ^ 2x + 2)?

Hvordan differentierer du sqrt (cos (x ^ 2 + 2)) + sqrt (cos ^ 2x + 2)?
Anonim

Svar:

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

Forklaring:

# (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) = (annuller2 (xsen (x ^ 2 + 2) + sen (x + 2))) / (annullér2sqrtcos (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)))