# Using Differentials to Approximate the Cube Root of 28

In this tutorial we shall look at the differentials of independent and dependent variables. Some applications of differentials will be discussed.

Use differentials to approximate the value of $\sqrt{{28}}$

The nearest number to 28 whose perfect cube root can be taken is 27, so let us consider that $x = 27$ and $\delta x = dx = 1$.

Now consider
$y = \sqrt{x} = {x^{\frac{1}{3}}}\,\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$
$\begin{gathered} y + \delta y = {\left( {x + \delta x} \right)^{\frac{1}{3}}} \\ \Rightarrow {\left( {x + \delta x} \right)^{\frac{1}{3}}} = y + \delta y \approx {\left( x \right)^{\frac{1}{3}}} + dx\,\,\,\,\,\,\,\therefore y = {\left( x \right)^{\frac{1}{3}}},\delta y \approx dy\,\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered}$

Taking the differential of equation (i), we have
$dy = d{\left( x \right)^{\frac{1}{3}}} = \frac{1}{3}{x^{\frac{1}{3} – 1}}dx = \frac{1}{3}{x^{ – \frac{2}{3}}}dx$

Putting this value in equation (ii), we have
$\begin{gathered} {\left( {x + \delta x} \right)^{\frac{1}{3}}} \approx {x^{\frac{1}{3}}} + \frac{1}{3}{x^{ – \frac{2}{3}}}dx \\ \Rightarrow {\left( {27 + 1} \right)^{\frac{1}{3}}} \approx {\left( {27} \right)^{\frac{1}{3}}} + \frac{1}{3}{\left( {27} \right)^{ – \frac{2}{3}}}\left( 1 \right)\,\,\,\,\,\,\,\because x = 27,\,dx = 1 = \delta x \\ \Rightarrow {\left( {28} \right)^{\frac{1}{3}}} \approx {\left( {{3^3}} \right)^{\frac{1}{3}}} + \frac{1}{3}{\left( {{3^3}} \right)^{ – \frac{2}{3}}} = 3 + \frac{1}{3}\left( {{3^{ – 2}}} \right) \\ \Rightarrow \sqrt{{28}} \approx 3 + \frac{1}{3} \times \frac{1}{9} = 3 + \frac{1}{{27}} = \frac{{81 + 1}}{{27}} = \frac{{82}}{{27}} = 3.037 \\ \end{gathered}$