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प्रश्न
Using differential, find the approximate value of `sqrt(36.6)` upto 2 decimal places.
बेरीज
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उत्तर
Suppose `f(x) = sqrt(x)`
⇒ f'(x) = `1/(2sqrt(x))`
We know that f'(x) = `(f(x + h) - f(x))/h`, where h → 0
Here, x = 36, h = 0.6
Then, f(36.6) = f(36) + 0.6f'(36)
⇒ `sqrt(36.6) = sqrt(36) + (0.6) xx 1/(2sqrt(36))`
⇒ `sqrt(36.6) = sqrt(36) + 6/10 xx 1/12`
⇒ `sqrt(36.6) = 6 + 1/20`
⇒ `sqrt(36.6) = 6 + 0.05`
⇒ `sqrt(36.6) = 6.05`
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