By making use of the relation \(\delta y \approx \dfrac{dy}{dx} \times \delta x\), and without using tables, prove that \(5.1\) is an approximate value of \(\sqrt{26}\).

Let’s consider the function \(y = \sqrt{x}\). We know that \(y(25) = \sqrt{25} = 5\) and that \[ \frac{dy}{dx} = y' = \frac{1}{2} x^{-1/2} \implies y'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{10}. \]

Now using \(\delta y \approx \dfrac{dy}{dx} \times \delta x\) at the point \(x = 25, y = 5\), with \(\delta x = 1\), we see that \(\delta y = 0.1 \times 1 \implies y + \delta y = 5.1.\)

Thus \(\sqrt{26} = \sqrt{25+1} \approx 5.1.\)

We can’t immediately say how accurate our approximation is here.

If we turn to our calculators, they tell us \(\sqrt{26} = 5.0990\!\ldots\), and so our approximation is surprisingly accurate (\(0.02\)% out).

With a problem like this, such accuracy cannot always be guaranteed!