What is the use of the diode in this circuit

Imagine the year is 1965, the Stones "I can't get no Satisfaction" is at the top of the charts, and we are using Vcc voltages of 10V or 15V for a discrete digital circuit. So the input voltage is between 0V and 15V. None of this 3.3V or 0.9V nonsense.

Without the diode, when the input is at 10V, the speedup capacitor will charge to about +9.3V (emitter is at ground, base is at +0.7V). Then when it goes low, the base voltage will be driven to about -9V (exceeding the typical -5V rating) and the BE junction will break down. Over time this will likely result in reduced beta and failure.