Matrix differentiation with square of trace leads to unpredictable zeros

In the Sympy live shell:

A = MatrixSymbol('A', 3,3)
(trace(Identity(3)-A)**2).diff(A)

delivers 0, with no error message.

Yet (trace(Identity(3)-A)**2).expand() delivers the right expansion.

In any case (trace(Identity(3)-A)**2).expand().diff(A) delivers 0.

The leading term (trace(A)**2) will, when explicitly typed, be correctly differentiated.