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Question
Using differentials, find the approximate value of `sqrt(0.082)`
Solution
Let f(x) = `sqrt(x)`
Using f(x + ∆x) = f(x) + ∆x . f′(x), taking x = 0.09 and ∆x = – 0.008,
We get f(0.09 – 0.008) = f(0.09) + (– 0.008) f′(0.09)
⇒ `sqrt(0.082) = sqrt(0.09) - 0.008 . (1/(2sqrt(0.09)))`
= `0.3 - (0.008)/0.6`
= 0.3 – 0.0133
= 0.2867.
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