[kalee20]

It is harder, yes.

Try this: Consider a capacitive anode load. Jerk up the grid by 1 volt and hold it steady.

Then, the anode current will immediately increase. Because the load is capacitive, the anode voltage will steadily ramp downwards.

So, Cga has a steadily ramping voltage across it (grid steady at +1V, anode ramping downwards). With a ramping voltage across this capacitance, a steady current will flow through Cga.

This current must flow out of the grid lead. Now, see what we have: A steady voltage change on the grid, and a steady current flowing at the grid. That is exactly what a resistor would do, so the grid 'looks' like a resistor. (The Cg(k+f+s) is still there, so it actually looks like this fictional resistor in series with this capacitance).

It's a bit harder to visualise the effect of an inductor. However, if you remember that an inductor gives a phase shift opposite to what a capacitor does, and a capacitive load gives current on one polarity through Cga, then an inductive load will give current of opposite polarity through Cga - so input looks like a negative resistance.