It does not make sense to do this. Consider the following chain of functions: y=C2(f(C1(x))):
x --> u=C1(x) --> v=f(u) --> y=C2(v)
Then computing the gradients of C1using the chain-rule relies on back-propagating the gradients of the output y all the way back to u. But if we cannot compute the gradients of function f , this chain is broken.