when taking partial derivatives of chiral superfileds respect to scalar components, which are the other variables?

Question

I am following some notes on supersymmetry and I need some clarification.

Chapter fice deals wuth sypersymmetric Lagrangians and the superpotential is introduced. It is stated that the interaction terms of chiral superfields are given by $$\mathcal{L}_{int}=\int{}d^2\theta\,W(\Phi)+h.c.$$ where $W$ is the superpotential and $\Phi$ is a chiral superfield. $$\Phi=\phi+\sqrt{2}\theta\psi+i\theta\sigma^{\mu}\bar{\theta}\partial_{\mu}\phi-\theta\theta{}F-\frac{i}{\sqrt{2}}\theta\theta\partial_{\mu}\psi\sigma^{\mu}\bar{\theta}-\frac{1}{4}\theta\theta\bar{\theta}\bar{\theta}\partial^2\phi$$ in page 81, it is said that this integral is easy if we notice that we can expand the superpotential $$W(\Phi)=W(\phi)+\sqrt{2}\frac{\partial{}W}{\partial\phi}\theta\psi-\theta\theta\left(\frac{\partial{}W}{\partial\phi}F+\frac{1}{2}\frac{\partial^2{}W}{\partial\phi\partial\phi}\psi\psi\right)$$ so $$\mathcal{L}_{int}=-\frac{\partial{}W}{\partial\phi}F-\frac{1}{2}\frac{\partial^2{}W}{\partial\phi\partial\phi}\psi\psi+h.c.$$ all this looks very good from afar. Nonetheless, let's take the specific case $$W=M\Phi^2$$ where $M$ is some mass constant and try to perform $$\frac{\partial{}W}{\partial\phi}=2M\Phi\frac{\partial\Phi}{\partial\phi}=2M\frac{\partial}{\partial\phi}\left(\phi+\sqrt{2}\theta\psi+i\theta\sigma^{\mu}\bar{\theta}\partial_{\mu}\phi-\theta\theta{}F-\frac{i}{\sqrt{2}}\theta\theta\partial_{\mu}\psi\sigma^{\mu}\bar{\theta}-\frac{1}{4}\theta\theta\bar{\theta}\bar{\theta}\partial^2\phi\right)$$ here goes my question, when taking the partial derivative of either $\Psi$ or $W$ respect to $\phi$, which are the other variables? Do I need to consider $\Psi(\phi,\psi,F)$? what do I do then with $\partial_{\mu}\phi$ and $\partial^2\phi$?

This post imported from StackExchange Physics at 2015-06-02 11:46 (UTC), posted by SE-user silvrfück