Using differentials, find the approximate value of `(1. 999)^5`

Explanation

Let `y =x^5`

and `x=2`

`x+trianglex=1.999`

`impliestrianglex=-0.001`

For `x=2,`

`y=(2)^5=32`

Let `dx=trianglex=-0.001`

As`y=x^5`,

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