1) f'(x) = 1/(2√x)
f'(x0) = 1/(2√4) = 1/(2*2) = 1/4
2) f'(x) = (sin(x))' + (1)' = cos(x) + 0 = cos(x)
f'(x0) = cos(π/3) = 1/2
3) f'(x) = (x^3)' + (6x)' - (2)' = 3x^(3-1) + 6 - 0 = 3x^2 + 6
f'(x0) = 3*(1^2) + 6 = 3*1 + 6 = 3+6=9
4) f'(x) = (√x)' + (1/x)' + (1)' = 1/(2√x) - 1/(x^2) + 0 = 1/(2√x) + 1/(x^2)
f'(x0) = 1/(2√4) - 1/(4^2) = 1/(2*2) - 1/16 = 1/4 - 1/16 = 4/16 - 1/16 = 3/16