Using differentials, find the approximate value of $\sqrt{51}$
$\sqrt{51}=\left(51\right)^{\frac{1}{2}}$
Let $y = x^{\frac{1}{2}}$
Let $x = 49$, $\Delta x = 2$ so that $x + \Delta x = 51$
Now, $\Delta y = \left(x + \Delta x\right)^{\frac{1}{2}} - x^{\frac{1}{2}}$
i.e. $\Delta y = \left(51\right)^{\frac{1}{2}} - \left(49\right)^{\frac{1}{2}}= \left(51\right)^{\frac{1}{2}}-7$
$\therefore$ $\left(51\right)^{\frac{1}{2}} = \Delta y + 7$ $\;\; \cdots$ (1)
Also $\dfrac{dy}{dx}= \dfrac{1}{2 \sqrt{x}}$
Now, $\Delta y$ is approximately equal to dy and
$dy = \left(\dfrac{dy}{dx}\right) \Delta x = \dfrac{1}{2 \sqrt{x}} \times \Delta x = \dfrac{1}{2 \sqrt{49}} \times 2 = \dfrac{1}{7}= 0.14286$
$\therefore$ Approximate value of $\Delta y = dy = 0.14286$
$\therefore$ We have from equation (1)
$\sqrt{51} = 0.14286 + 7 = 7.14286$