Question

1. y = (1-2x)^1/2
2. y = sin^3 (x-2)
3 y = cos^5 (3-2x)

Answer 1

Yang diminta turunannya ya
1. y = (1-2x)^1/2
    y' = (1/2)(1 - 2x)^(-1/2)(-2)
       = -(1 - 2x)^(-1/2)
       =     - 1        
           √(1 - 2x)
2. y = sin^3 (x-2) = (sin(x-2))³
   y'  = 3(sin(x-2))²(cos(x-2))(1)
       = 3sin²(x-2)cos(x-2) atau

3 y = cos^5 (3-2x) = (cos(3-2x))^5
   y' = 5(cos(3-2x))^4(-sin(3-2x))(-2)
       = 10cos^4(3-2x)sin(3-2x)