Use Differentials to Approximate Square Root of 49.5

In this tutorial we shall develop the differentials 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)


Differentiate 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


Putting 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}