A one dimensional complex supermanifold $X$ is locally described by an ordinary complex coordinate $z$ and an anticommuting coordinate $\theta$, $\theta^2 = 0$.


The superderivative is the square root of the derivative in the following sense:
Take the vector field $D\_{\theta} = \partial\_{\theta} + \theta\partial\_z$ so that $\frac12[D\_{\theta},D\_{\theta}]= \partial\_z$, where $[ \ , \ ]$ denotes the super-commutator.


It is stated (see for example pg. 4 Equation 2.9 in <https://arxiv.org/pdf/1209.2459.pdf>) that under a superanalytic change of coordinates $(z, \theta) \mapsto (\overline{z}(z, \theta), \overline{\theta}(z, \theta))$ the superderivative $D\_{\theta}$ transforms as
$$ D\_{\theta} = (D\_{\theta} \overline{z} + \overline{\theta} D\_{\theta} \overline{\theta} )\frac{\partial}{\partial\_{\overline{z}}} + D\_{\theta} \overline{\theta} D\_{\overline{\theta}} $$


*My Question:* When I use the chain rule I find that:
$$D\_{\theta} = (D\_{\theta} \overline{z})\frac{\partial}{\partial\_{\overline{z}}} + D\_{\theta} \overline{\theta} D\_{\overline{\theta}}. $$


In particular, I am missing the term $\overline{\theta} D\_{\theta} \overline{\theta} $ term. I cannot figure out how to reproduce the given transformation $D\_{\theta}$ using the ordinary chain rule from calculus.


How does one use the chain rule on $D\_{\theta}$ to produce the term $\overline{\theta} D\_{\theta} \overline{\theta} $?



Chain rule for the superderivative
Direct application of definitions and chain rule gives
$$
D\_\theta = (D\_\theta \hat{\theta}) \partial\_{\hat{\theta}}+(D\_\theta \hat{z}) \partial\_{\hat{z}}.
$$
Then eliminate $\partial\_{\hat{\theta}}$ in favour of $D\_{\hat{\theta}}$:
$$
D\_\theta = (D\_\theta \hat{\theta}) (D\_{\hat{\theta}}-\hat{\theta}\partial\_{\hat{z}})+(D\_\theta \hat{z}) \partial\_{\hat{z}}.
$$
Rearranging terms, we get
$$ D\_\theta =(D\_\theta \hat{\theta})D\_{\hat{\theta}} +(D\_\theta \hat{z}-\hat{\theta}D\_\theta \hat{\theta}) \partial\_{\hat{z}}
$$
as claimed in (2.9) of the given reference (note the sign).


