# $R$-Symmetry of gauge field

Suppose $$V$$ is a superfield scalar under R-transformations. This means that under an R-transformation $$V\mapsto V'$$ where $$V'(x,\theta,\bar{\theta})=V(x,e^{-iK}\theta,e^{iK}\bar{\theta})$$. What is then the transformation of $$W_\alpha:=-\frac{1}{4}\bar{D}_{\dot{\alpha}}\bar{D}^{\dot{\alpha}}D_\alpha V$$, where $$D_\alpha f(x,\theta,\bar{\theta}):=\frac{\partial f}{\partial\theta^\alpha}(x,\theta,\bar{\theta})+i\sigma^m_{\alpha\dot{\alpha}}\bar{\theta}^\dot{\alpha}\frac{\partial f}{\partial x^m}(x,\theta,\bar{\theta}),$$ and $$\bar{D}^\alpha f(x,\theta,\bar{\theta}):=\frac{\partial f}{\partial\bar{\theta}_\dot{\alpha}}(x,\theta,\bar{\theta})+i\bar{\sigma}^m_{\dot{\alpha}\alpha}\theta^\alpha\frac{\partial f}{\partial x^m}(x,\theta,\bar{\theta}),$$ for all superfields $$f$$? I would imagine $$W_\alpha\mapsto W_\alpha'$$ where $$W_\alpha'(x,\theta,\bar{\theta})=-\frac{1}{4}\bar{D}_{\dot{\alpha}}\bar{D}^{\dot{\alpha}}D_\alpha V'(x,\theta,\bar{\theta}).$$ I would however like to show that this is precisely $$W_\alpha'(x,\theta,\bar{\theta})=e^{iK}W_\alpha(x,e^{-iK}\theta,e^{iK}\bar{\theta}).$$ I need some help with this simple vector calculus exercise. Apparently I don't know my chain rule.

Note that $$D_{\alpha}$$ and $$D_{\dot\alpha}$$ transform under the R-symmetry as
$$D_{\alpha}\rightarrow e^{-iK}D_{\alpha},\qquad \bar D_{\dot\alpha}\rightarrow e^{+iK}\bar D_{\dot\alpha}$$
so we get $$W_{\alpha}\rightarrow e^{+iK}W_{\alpha}$$. This is analogous to how usual derivatives transforms under scaling $$x\rightarrow \Lambda x$$:
$$\frac{d}{dx}\rightarrow \frac{d}{d(\Lambda x)}=\Lambda^{-1}\frac{d}{dx}$$
or more general, a differomorphism $$x\rightarrow y(x)$$:
$$\frac{d}{dx}\rightarrow \frac{d}{dy}=(\frac{dx}{dy})\frac{d}{dx}$$