Let x be the edge of the cube, V be the volume of the cube at any instant of the time.

We know,

V = x³

Differentiating the above equation with respect to time, we get

But is is given that the volume of the cube is increasing at the rate of 9cm³/sec, so the above equation becomes,

We also know that the surface area of the cube is

S = 6x²

Again differentiating the above equation with respect to time is

Substitute equation (ii) in above equation, we get

When the edge is of length 10 cm, we get

Hence the rate at which the surface area increasing when the length of an edge is 10 cm is 3.6 cm²/sec