If the original 24 m edge length x of a cube decreases at the rate 5m/min, when x=3 at what rate does the cube's surface area change

Let s be the surface area and t = time in minutes

s=6x²

d/dt (s) = d/dt (6x²)

ds/dt=12x (dx/dt)

given that x=3, then

ds/dt=12 (3) (-5)

ds/dt=-180m²/min

b] Let the volume be v

v=x³

d/dt (v) = d/dt (x³)

dv/dt=3x² (dx/dt)

when x=3, then

dv/dt=3 (3²) (-5)

dv/dt=-135m³/min