Using differentials, find the approximate value of `sqrt(26)`

Explanation

Let `y=sqrtx=x^{1/2}`

and `x=25`

`x+trianglex=26`

`impliestrianglex=1`

For `x=25,`

`y=(25)^{1/2}=5`

Let `dx=trianglex=1`

As`y=x^{1/2}`,

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