Cos (1-2x) ^ 2를 어떻게 구별합니까?

대답:

# dy / dx = 4cos (1-2x) sin (1-2x) #

설명:

첫째, #cos (1-2x) = u #

그래서, # y = u ^ 2 #

# dy / dx = (dy) / (du) * (du) / (dx) #

# (dy) / (du) = 2u #

(dx) = d / dx cos (1-2x) = d / dx cos (v) #

# (du) / (dx) = (du) / (dv) * (dv) / (dx) #

# dy / dx = (dy) / (du) * (du) / (dv) * (dv) / (dx) #

# (du) / (dv) = - sin (v) #

# (dv) / (dx) = - 2 #

# dy / dx = 2u * -sin (v) * - 2 #

# dy / dx = 4usin (v) #

# dy / dx = 4cos (1-2x) sin (1-2x) #