Use differentials to approximate the cube root of 127.

Explanation

`y = f(x) = (x)^(1/2)`

`(127)^(1/3) = f(x + /_ x) `

when `x=125 & /_x=2`

`f(x+ /_ x) = /_ y + f(x) `

`f(x) = f(125) = (125)^(1/3) = 5`

`/_ y = dy = (dy/dx) /_/x`

`= d/dx(x^(1/3)) xx 2`

`= [1/(3x^(2/3))] xx 2`

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