Using differentials, find the approximate value of `sqrt(25. 02)`

Explanation

`y= f(x) = sqrtx`

`y = f(x +Delta x) - f(x)`

`x= 25 & Delta x= 0.2`

`=> f(x + Delta x) = f(x) + Delta y`

`f(x) = f(25) = 5`

`Deltay = dy = (dy/dx)Deltax = (d(sqrtx))/(dx) xx 0.2 `

`Delta y = 1/(2 sqrtx) xx 0.2`

`= 0.1/5 = 1/50`

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