Solving circuit DEs: derivative of inductor current should be zero, but isn't
[ last line should be i'L(0+) ]
Your calculations are correct. At t=0+, there is 50 mA in the inductor, but 0 V across it (and the capacitor). So i'(L) =0. Now, the voltage across the capacitor will increase (because supply increased from 5 to 10 V; the 'extra' 50 mA will flow in the capacitor. The cap voltage will rise, and then the inductor current will increase.

In a small region around 0+, the cap V will increase linearly, and inductor current quadratically with time, however overall the result will be (exponentially decaying) sinusoids. 

Clearly at the extreme of a sinusoid, the derivative is 0, yet since the 2nd derivative isn't 0, the derivative will change, and so will the value.