alleryFind the approximate value of $\sqrt{25.5}$ using differentials (given $\sqrt{25} = 5$):
A5.05
B5.5
C5.0498 (more precise)
D5.1
Answer & Solution
Correct answer: A. 5.05
Let $f(x) = \sqrt x$. $f'(x) = 1/(2\sqrt x)$.
$\Delta f \approx f'(25)\cdot 0.5 = (1/10)(0.5) = 0.05$.
**So $\sqrt{25.5} \approx 5 + 0.05 = 5.05$.**
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