Kako razlikujete cos (1-2x) ^ 2?

Kako razlikujete cos (1-2x) ^ 2?
Anonim

Odgovor:

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

Pojasnilo:

Najprej pustite #cos (1-2x) = u #

Torej, # y = u ^ 2 #

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

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

# (du) / (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) #