The approximate value of $${\left( {0.007} \right)^{\frac{1}{3}}} = ?$$

Solution :
$$\eqalign{ & {\text{Let }}f\left( x \right) = {x^{\frac{1}{3}}}\,\, \Rightarrow f'\left( x \right) = \frac{1}{3}{x^{ - \frac{2}{3}}} \cr & {\text{Now, }}f\left( {x + \Delta x} \right) - f\left( x \right) = f'\left( x \right).\Delta x = \frac{{\Delta x}}{{3\left( {{x^{\frac{2}{3}}}} \right)}} \cr & {\text{We may write, }}0.007 = 0.008 - 0.001, \cr & {\text{taking }}x = 0.008{\text{ and }}dx = - 0.001 \cr & {\text{we have, }}f\left( {0.007} \right) - f\left( {0.008} \right) = - \frac{{0.001}}{{3{{\left( {0.008} \right)}^{\frac{2}{3}}}}} \cr & \Rightarrow f\left( {0.007} \right) - {\left( {0.008} \right)^{\frac{1}{3}}} = - \frac{{0.001}}{{3{{\left( {0.2} \right)}^2}}} \cr & \Rightarrow f\left( {0.007} \right) = 0.2 - \frac{{0.001}}{{3\left( {0.04} \right)}} \cr & \Rightarrow f\left( {0.007} \right) = 0.2 - \frac{1}{{120}} \cr & \Rightarrow f\left( {0.007} \right) = \frac{{23}}{{120}} \cr & {\text{Hence, }}{\left( {0.007} \right)^{\frac{1}{3}}} = \frac{{23}}{{120}} \cr} $$