The volume of a cube is increasing at the rate of $\displaystyle 6 \mathrm{~cm}^{3} / \mathrm{s}$. How fast is the surface area of cube increasing, when the length of an edge is $\displaystyle 8$ cm ?

Given, $\displaystyle \frac{d V}{d t}=6 \mathrm{~cm}^{3} / \mathrm{sec}$. Since, $\displaystyle V=x^{3}$ $\displaystyle \frac{d V}{d t}=3 x^{2} \frac{d x}{d t} \Rightarrow 6=3 x^{2} \frac{d x}{d t} \Rightarrow \frac{d x}{d t}=\frac{2}{x^{2}} \mathrm{~cm} / \mathrm{sec}$ Now, Surface Area $\displaystyle =\mathrm{S}=6 x^{2} \Rightarrow \frac{d \mathrm{S}}{d t}=12 x \frac{d x}{d t}=3 \mathrm{~cm}^{2} / \mathrm{sec}$