Use differentials to approximate the cube root of 127.

Explanation

Let `y = root(3)x = x^(1//3)`
Let x = 125 and x + ` Deltax = 127`
` rArr Deltax = 2`
` At x = 125`
`y = root(3)125 =5`
Now `y= x^(1//3)`
`rArr (dy)/(dx) = 1/3 x^(-2//3) = 1/ (3x^(2//3))`
At x = 125,
` (dy)/(dx) = 1/(3(125)^(2//3)) = 1/75`
`:. Delta y = (dy)/(dx) . Deltax = 1/75 xx 2 = 2/75 = 0.026`
` and y+Deltay = 5 + 0.026`
` rArr (127)^(1//3)= 5.026`.