The volume of a cube is increasing at $9$ cm$^3$/s. When the edge length is $10$ cm, the surface area is increasing at the rate of

1. $V=x^3$, $S=6x^2$. 2. $\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}=9 \Rightarrow \dfrac{dx}{dt}=\dfrac{3}{x^2}$. 3. $\dfrac{dS}{dt}=12x\dfrac{dx}{dt}=12x\cdot\dfrac{3}{x^2}=\dfrac{36}{x}$. 4. At $x=10$: $\dfrac{dS}{dt}=\dfrac{36}{10}=3.6$. 5. The trap $7.2$ wrongly uses $S=3x^2$. _Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.1_