What happpens to the Vgs in the Miller plateau region during MOSFET turn-ON?
Here's a typical MOSFET switching circuit, with one component, normally hidden inside the MOSFET, here explicitly drawn on the circuit as C1.

There are actually capacitances between each pair of electrodes, but Cdg (=C1 here) is especially important in this configuration, so we can (to a first approximation) ignore the others (Cgs and Cds). 

The reason for its importance comes from the relatively large change in Vds caused by a small change in Vgs, which causes a relatively large current through C1 opposing the current supplied by the driver. This has been known as the Miller effect, and Cdg as the Miller capacitance, since at least the early 1930s.







simulate this circuit &ndash; Schematic created using CircuitLab

So consider what happens when V1 switches from 0V to 5V.


The MOSFET is initially switched off, so Vds=100V. Vgs increases as the gate capacitances Cgs and Cdg charge in parallel through R1. You can estimate these capacitances from the slope at the start of the Vgs curve.
When Vgs passes the threshold voltage, M1 starts to turn on, and Vds starts to fall. Now there is a current through Miller capacitance C1 (Cdg), almost equal and opposite to that through R1, so Vgs remains flat (the plateau). I=C1* dVdg/dT
When the transistor is fully on (saturated), Vds=0 so this current ceases, and Vgs increases to 5V as Cgs and Cdg complete charging.


Much the same happens on switching off, so a good MOSFET driver must both source and sink current, probably better than the 100 ohm resistor R1 on the drawing.

Check the rise times on a datasheet's Vgs curves against the datasheet capacitance values and driver impedances, they should all make reasonable sense given the above explanation.