1) d/dx(arctan(1-x) = -1/(1-(1-x)^2 = -1/sqrt(2x-x^2)

2) tan(arcos(x)) = sqrt(1-x^2)/x draw a right triangle with angle whose cosine is x

dsqrt(1-x^2)/x /dx = -1/sqrt(1-x^2) - sqrt(1-x^2)/x^2

= -1/[x^2 sqrt(1-x^2)]

Even without this d(tan(u)/dx = sec^2 (u)8du/dx

d(arctan(cos(x))/dx = sec^2 (cos(x)) [-1/sqrt(1-x^2)]

sec(arccos(x)) = 1/x = -1/[x^2 sqrt(1-x^2)]

for 3) 9x^2 -6x +8 = (3x-1)^2 + 7 let u =3x-1 and note a = sqrt(7)