Find the approximate change in the value V of a cube of side x metres caused by increasing the side by 1%.

Volume of the cube, $V=x^{3}$

We have

$\Delta x=0.01 x$

$\frac{d V}{d x}=3 x^{2}$

$\Rightarrow \Delta V=d V=\frac{d V}{d x} d x=3 x^{2} \times 0.01 x=0.03 x^{3}$

Hence, the approximate change in the value $V$ of the cube is $0.03 x^{3} \mathrm{~m}^{3}$.

Disclaimer: This solution has been created according to the question given in the book. However, the solution in the book is incorrect.