using differentials find the approimate vale of (0.999)1/10 ,upto three places of decimal

Let y = f(x) = x^0.1
Now dy/dx = f'(x) = 0.1x^-0.9
Again 
f(x) = f(x') + f'(x') (x - x')
let x' = 1 and x =0.999
Thus , 
f(0.999)  = f(1) + f'(1) ( -0.001)
f(0.999) = 1 - 0.001 * 0.1 * 1 = 1 - 0.0001 = 0.9999
thus , 0.999^1/10 = 0.9999