Use differentials to find the approximate value of `(log)_e(4. 01),` having given that `(log)_e4=1. 3863` .

Explanation

let `y= f(x) = log_e x`

`x + /_ x = 4.01`

`x + /_ x = 4 + 0.01`

`x= 4 & /_ x = 0.01`

`y= f(4) = log_e 4 = 1.3863`

`/_ x = 0.01`

`del x = 0.01`

`dx = 0.01`

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