You may go with polynomials over $\mathbf F_2$, the smallest finite field and one of the representations of Boolean functions. Note, that in $\mathbf F_2$ for any $x$ (i.e. $1$ or $0$) the following relations hold:
$x^2 = x, \quad x + x = 0.$
The derivative is defined as:
$\left(D \, f \right)(x) = f(x+1) - f(x)$
which is also a finite difference. All the usual derivative properties, including the chain rule, are fulfilled. Indeed one can easily show it by studying every possible $f$ --- all the two of them: identity $x \mapsto x$ and negation $x \mapsto x+1$.
Well, considering one-variable functions $\mathbf F_2 \to \mathbf F_2$ is not very interesting and it is better to proceed with multivariable calculus on $\mathbf F_2^N = \underbrace{\mathbf F_2 \times \ldots \times \mathbf F_2}_N $. The directional derivative is defined as
$\left(D_\boldsymbol u \, g \right)(\boldsymbol x) = g(\boldsymbol x+ \boldsymbol u) + g(\boldsymbol x)$
where
$g : \mathbf F_2^N \to \mathbf F_2, \quad \boldsymbol x, \boldsymbol u : \mathbf F_2^N$
and $+$ is defined as component-wise addition:
$\boldsymbol x + \boldsymbol u = \left(x_1 + u_1, \ldots , x_N + u_N \right).$
The chain rule for the above g and $f : \mathbf F_2 \to \mathbf F_2$ is fulfilled:
$\left(D_\boldsymbol u \; f \circ g \right)(\boldsymbol x) = (D_\boldsymbol u \; f) \, (g(\boldsymbol x)) \cdot (D_\boldsymbol u g)(\boldsymbol x),$
which again is possible to prove by considering every possible $f$. At least this is how I was able to prove it, and I am not in anyway a mathematician. Unfortunately, a quick search over Google books reveals little (nothing?) on the chain rule for boolean functions. As for me I used "Boolean Functions in Coding Theory and Cryptography" by O. A. Loginov, A. A. Salnikov, and V. V. Yashchenko for the introduction to boolean functions, but again it does not go into the chain rule.
I would appreciate if anyone would point me some books or articles on the topic of derivatives of boolean functions, especially when one deals with functions of the type $\mathbf F_2^N \to \mathbf F_2^M$.