\(\sqrt {10} \) is near \(\sqrt 9 \), so we will use \(f\left( x \right) = \sqrt x \) with x = 9 and Δx = 1. Note that \(f'\left( x \right) = \frac{1}{{2\sqrt x }}\).

\(\eqalign{ \sqrt {10} &= f\left( {x + \Delta x} \right)\\ &\approx f\left( x \right) + f'\left( x \right)\Delta x\\ &= \sqrt x + \frac{1}{{2\sqrt x }}\Delta x\\ &= \sqrt 9 + \frac{1}{{2\sqrt 9 }}\left( 1 \right)\\ &= 3\frac{1}{6} }\) .

Thus we see that \(\sqrt {10} \approx 3\frac{1}{6} = 3.166\bar 6\). This is very close to the correct value of \(\sqrt {10} \approx 3.1623\).