At time t = 0, a storage tank is empty and begins filling with water. For t > 0 hours, the depth of the water in the tank is increasing?

Correct answers:

Two hours after the tank begins filling with water, W is increasing at a rate greater than 3 feet per hour, per hour.

Two hours after the tank begins filling with water, the rate at which the depth of water is rising is increasing at a rate greater than 3 feet per hour, per hour.

It is like option c) except for the last two words. None of the options you present are completely correct.

Of course, W(2) > 3 (without the derivative) would be option (a) 

The reason W ' (t) is measured in feet per hour per hour (ft / h^2) is because the rate of increase of depth (speed, W) is ALREADY measured in ft / h. You cannot also use ft / h to measure the rate of change of it w.r.t. time.

Basically, height ---> feet

W = speed ---> feet / hour

W ' (t) = acceleration ---> (feet / hour) / hour

So suppose at t=5, W = 20 ft/h

And at t=6, W = 30 ft/h.

The "average rate of change" of W from t=5 to t=6 would be 10 ft/h^2 because it literally increased by 10 feet per hour in one hour.

The derivative W ' (t) is the instantaneous rate of change rather than average, but the procedure is anologous.