letf(x) = (1+ x)^(1/3) =>f(0)=1f'(x) =(1/3)(1+x)^(-2/3) =>f'(0) = 1/3(1+x)^(1/3) ~ f(0) + f'(0) x = 1 + (1/3)x(1+x^2)^(1/3) ~ 1 + (1/3)x^2(1+x^2)^(1/3) - 1~ (1/3)x^2
