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Using differentials, find the approximate value of 0.082

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Question

Using differentials, find the approximate value of `sqrt(0.082)`

Sum
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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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Chapter 6: Application Of Derivatives - Solved Examples [Page 122]

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NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 6 Application Of Derivatives
Solved Examples | Q 7 | Page 122
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