Verified answer

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

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

dy/dx = d/dx (5x) (1-x)^(-2/3) + 5x d/dx ((1-x)^(-2/3)) + d/dx (cos^2 (2x+1))

d/dx (5x) = 5

d/dx ((1-x)^(-2/3)) = (-2/3) (1-x)^(-5/3) d/dx ((1-x)) = (-2/3) (1-x)^(-5/3)(-1) = (2/3) (1-x)^(-5/3)

d/dx (cos^2 (2x+1)) = 2 cos(2x+1) d/dx (cos(2x+1) = 2 cos(2x+1) (-sin(2x+1)) d/dx (2x+1) = -2 cos(2x+1)sin(2x+1)(2) = -4 cos(2x+1)sin(2x+1)

dy/dx = 5 (1-x)^(-2/3) + 5x (2/3) (1-x)^(-5/3) -4 cos(2x+1)sin(2x+1)

dy/dx = 5/(1-x)^(2/3) + 10x/ (3(1-x)^(5/3)) - 4 cos(2x+1)sin(2x+1)

y = 5x/[(1 - x)^(2/3)] + [cos(2x+1)]^2

To find A(x) = d/dx of 5x/[(1 - x)^(2/3)], apply the quotient rule

d/dx[u(x)/v(x)] = (vu – uv )/v^2

A = [(1 - x)^(2/3) *5 – 5x*(2/3)(-1)/ (1 - x)^(1/3)]/[(1 - x)^(4/3)]

Multiply top and bottom by 3(1 - x)^(1/3)

A = [15 – 15x + 10x*1] /[3(1 - x)^(5/3)] = 5(3 – x) /[3(1 - x)^(5/3)]… (1)

Let B(x) = d/dx of [cos(2x+1)]^2 = 2* cos(2x+1)* -sin(2x+1)*2 …………(2)

dy/dx = A + B = 5(x – 3)/ /[(x - 1)^(5/3)] – 2 sin(4x + 2)