In Chapter 1 of Nakahara's Geometry, Topology, Physics, Grassman numbers are defined as linear combinations of objects $\theta_i$ which satisfy anti-commutation relations $\{ \theta_i, \theta_j\} = 0$. Then differentiation with respect to these $\theta_i$ variables is introduced, and is stated to follow these two rules:
$\frac{d}{d\theta_i}\theta_j = \delta_{ij}$
"It is also assumed that the differential operator anti-commutes with $\theta_k$." My reading of this is that for any given element $f$ of the Grassman algebra, the following holds: $\frac{d}{d\theta_i}(\theta_k\times f) + \theta_k \times \frac{d}{d\theta_i} f = 0$
Nakahara then writes "The Leibnitz rule then takes the form
$\frac{d}{d\theta_i}(\theta_j \theta_k) = \frac{d\theta_j}{d\theta_i}\theta_k - \theta_j\frac{d\theta_k}{d\theta_i}$,"
but I don't see how this follows from the previous two rules. Naively, I would think that I could evaluate $\frac{d}{d\theta_i}(\theta_j \theta_k)$ by letting $\theta_k$ play the role of the function $f$ in rule (2) above, and letting $\theta_j$ play the role of $\theta_k$. Explicitly, from rule (2) we know that
$\frac{d}{d\theta_i}(\theta_k\times f) + \theta_k \times \frac{d}{d\theta_i} f = 0$
Then making the substitution $j\rightarrow k$ gives
$\frac{d}{d\theta_i}(\theta_j\times f) + \theta_j \times \frac{d}{d\theta_i} f = 0$
Then since $f$ is any function, let $f = \theta_k$, so we have
$\frac{d}{d\theta_i}(\theta_j\times \theta_k) + \theta_j \times \frac{d}{d\theta_i} \theta_k = 0$
So
$\frac{d}{d\theta_i}(\theta_j\times \theta_k) = - \theta_j \times \frac{d}{d\theta_i} \theta_k$
Please help me understand where that derivation goes wrong. Thanks!
