# Using Differentials to Approximate the Square Root of 49.5

In this tutorial we shall develop the differentials to approximate the value of $\sqrt {49.5}$.

The nearest number to 49.5 whose square root can be taken is 49, so let us consider that $x = 49$ and $\delta x = dx = 0.5$.

Now consider
$y = \sqrt x \,\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Differentiating equation (i) with respect to $x$, we have
$\begin{gathered} \frac{{dy}}{{dx}} = \frac{d}{{dx}}\sqrt x \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }}\,\,\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered}$

Taking the differential of equation (ii), we get
$\Rightarrow dy = \frac{1}{{2\sqrt x }}dx$

Using the values $x = 49$ and $dx = 0.5$, we have
$\begin{gathered} dy = \frac{1}{{2\sqrt {49} }}\left( {0.5} \right) = \frac{{0.5}}{{2\left( 7 \right)}} \\ \Rightarrow dy = 0.0357 \\ \end{gathered}$

Now
$\begin{gathered} \sqrt {49.5} = y + dy \\ \Rightarrow \sqrt {49.5} = \sqrt x + dy \\ \Rightarrow \sqrt {49.5} = \sqrt {49} + 0.0357 = 7 + 0.0357 \\ \Rightarrow \sqrt {49.5} = 7.0357 \\ \end{gathered}$