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Using differentials, find the approximate value of $\sqrt{ 25.3}$ up to two places of decimals.

Answers (1)

Assume

$f(x) =\sqrt{x}$

$f'(x) =\frac{1}{2\sqrt{x}}$

$f(x+\Delta x) =f(x)+\Delta x f'(x)$

$f(25.3) =f(25)+ 0.3 f'(25)$

$\sqrt{25.3}= f(25.3) =5+ 0.3 \times \frac{1}{2 \times 5} = 5.03$

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