The side length of a cube is increasing at a constant rate of 2 cm per second. Using differentiation, which of the following is the rate of increase of the cube's total surface area, in cm² per second, at the instant when the side length is 5 cm?

AS Level Maths 9709: Connected Rates of Change for a Cube's Surface Area

A cube has $6$ faces, each of area $x^2$ , so the total surface area is

$S = 6x^2$

$\frac{dS}{dx} = 12x$

We are given $\dfrac{dx}{dt} = 2$ (constant). By the chain rule,

$\frac{dS}{dt} = \frac{dS}{dx} \times \frac{dx}{dt} = 12x \times 2 = 24x$

$\frac{dS}{dt} = 24(5) = 120$

Working directly at $x=5$ : $\dfrac{dS}{dx}\big|_{x=5} = 12(5) = 60$ . Then $\dfrac{dS}{dt} = 60 \times 2 = 120$ , matching Step 4 exactly.

A ( $60$ ): this is $\dfrac{dS}{dx}$ at $x=5$ , but the rate $\dfrac{dx}{dt}=2$ was never multiplied in.
B ( $20$ ): comes from differentiating a single face’s area $x^2$ instead of the total surface area $6x^2$ , giving $\dfrac{dS}{dx}=2x=10$ at $x=5$ , then $10\times2=20$ .
C ( $150$ ): comes from confusing surface area with volume, using $V=x^3 \to \dfrac{dV}{dx}=3x^2=75$ at $x=5$ , then $75\times2=150$ .

$\boxed{\frac{dS}{dt} = 120 \text{ cm}^2\text{/s}} \quad \text{(Option D)}$