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