Using differentials, find the approxi...

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

Text Solution

Verified by Experts

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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