The approximate value of ${(8.01)}^{\frac{4}{3}}+{(8.01)}^2$ upto 3 decimal places is

$\begin{array}{rcl}\mathrm{x}& =& 8 \mathrm{\delta x}=0.01\\ \mathrm{f}\left(\mathrm{x}\right)& =& {\mathrm{x}}^{4/3}+{\mathrm{x}}^2\\ \mathrm{f}(\mathrm{x}+\mathrm{\delta x})& =& {\mathrm{f}}^1\left(\mathrm{x}\right).\mathrm{\delta x}+\mathrm{f}\left(\mathrm{x}\right)\\ & =& \left(\frac{4}{3} {\mathrm{x}}^{1/3}+2\mathrm{x}\right)\mathrm{\delta x}+\mathrm{f}\left(\mathrm{x}\right)\\ & =& \left(\frac{8}{3}+16\right)(0.01)+(16+64)\\ & =& 80.186\end{array}$