Given a cube whose side x is decreased by 1%.

Let Δx be the change in the value of x.

Hence, we have

∴ Δx = –0.01x

The surface area of a cube of radius x is given by

S = 6x²

On differentiating A with respect to x, we get

We know

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = –0.01x

⇒ ΔS = (12x)(–0.01x)

∴ ΔS = –0.12x²

Thus, the approximate change in the surface area of the cube is 0.12x² m².