Using differentials, find the approximate value of `sqrt(401)`

Explanation

We have to find the approximate value of
`sqrt401 =
sqrt(400+1)`

​Let` y=f(x)= sqrtx`

`​y+Δy= sqrt(x+Δx)`

​`⇒Δy= sqrt(x+Δx)- sqrtx`

​Also,`Δy=f' (x)Δx`

`⇒ sqrt(x+Δx)− sqrtx=1/(2 sqrtx)Deltax`
Put `x=400,Δx=1`

`⇒ sqrt401 − sqrt400=1/40`
`=> sqrt401
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