Use differentials to approximate the cube root of 127.

Explanation

Let `f(x) = x^( ⅓)`. Then , `f(x) = (1)/(3x^(⅔))`
Now, `{f (x + deltax) - f(x)} = f'(x).deltax`
`rArr {f(x + deltax) - f(x)) = (1)/(3x^(⅔)).deltax`...(i)
We may write, 127 = (125 +2)
Putting x = 125 and `delta x = 1` in (i), we get
`f(125 + 2) -f(125) = (1)/(3 xx (125)^(⅔)) xx 2`
`rArr f(127) - f(125) = (2)/(75)`
`rArr f(127) = f(125) + (2)/(75) = {(125)^(⅓) + (2)/(75)} = (5 + (2)/(75)) = (377)/(75)`
`rArrroot(3)(127) = (377)/(75) = 5.026`