2. (4) At a certain instant each edge of a cube is $5 \mathrm{in}$. long and the volume is increasing at the rate of $2 \mathrm{in}^{3} / \mathrm{min}_{\text {. }}$ How fast is the surface area of the cube increasing?

Step 1 ：We are given a cube with edge length $s = 5$ inches and the rate of change of volume $\frac{dV}{dt} = 2$ cubic inches per minute. We are asked to find the rate of change of the surface area $\frac{dA}{dt}$.

Step 2 ：The volume $V$ of a cube is given by $V = s^3$ and the surface area $A$ is given by $A = 6s^2$.

Step 3 ：We can use the chain rule to relate these quantities. The chain rule states that if a variable $z$ depends on the variable $y$ which itself depends on the variable $x$, then $\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}$.

Step 4 ：In this case, the surface area $A$ depends on the side length $s$ which itself depends on time $t$. So we can write $\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt}$.

Step 5 ：We can find $\frac{dA}{ds}$ by differentiating the formula for the surface area with respect to $s$, which gives $\frac{dA}{ds} = 12s$.

Step 6 ：We can find $\frac{ds}{dt}$ by differentiating the formula for the volume with respect to $t$ and rearranging to solve for $\frac{ds}{dt}$. This gives $\frac{ds}{dt} = \frac{1}{3s^2} \cdot \frac{dV}{dt}$.

Step 7 ：Substituting the given values $s = 5$ and $\frac{dV}{dt} = 2$ into the formula for $\frac{ds}{dt}$ gives $\frac{ds}{dt} = \frac{2}{75}$.

Step 8 ：Finally, substituting $s = 5$ and $\frac{ds}{dt} = \frac{2}{75}$ into the formula for $\frac{dA}{dt}$ gives $\frac{dA}{dt} = 12 \cdot 5 \cdot \frac{2}{75} = 2$ square inches per minute.

Step 9 ：\(\boxed{\frac{dA}{dt} = 2 \, \text{in}^2/\text{min}}\)