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

Let Δx be the change in the value of x.

Hence, we have

∴ Δx = 0.01x

The volume of a cube of radius x is given by

V = x³

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

⇒ ΔV = (3x²)(0.01x)

∴ ΔV = 0.03x³

Thus, the approximate change in the volume of the cube is 0.03x³ m³.