Partial derivatives of chiral superfields with respect to scalar components in the superpotential I am following some notes on supersymmetry by Matteo Bertolini and I need some clarification.
Chapter five deals with sypersymmetric Lagrangians and the superpotential is introduced. It is stated that the interaction terms of chiral superfields are given by
$$\mathcal{L}_{int}=\int{}d^2\theta\,W(\Phi)+h.c.\tag{5.15}$$
where $W$ is the superpotential and $\Phi$ is a chiral superfield. 
$$\Phi=\phi+\sqrt{2}\theta\psi+i\theta\sigma^{\mu}\bar{\theta}\partial_{\mu}\phi-\theta\theta{}F-\frac{i}{\sqrt{2}}\theta\theta\partial_{\mu}\psi\sigma^{\mu}\bar{\theta}-\frac{1}{4}\theta\theta\bar{\theta}\bar{\theta}\partial^2\phi\tag{4.54}$$
in page 81, it is said that this integral is easy if we notice that we can expand the superpotential
$$W(\Phi)=W(\phi)+\sqrt{2}\frac{\partial{}W}{\partial\phi}\theta\psi-\theta\theta\left(\frac{\partial{}W}{\partial\phi}F+\frac{1}{2}\frac{\partial^2{}W}{\partial\phi\partial\phi}\psi\psi\right)\tag{5.19}$$
so
$$\mathcal{L}_{int}=-\frac{\partial{}W}{\partial\phi}F-\frac{1}{2}\frac{\partial^2{}W}{\partial\phi\partial\phi}\psi\psi+h.c.\tag{5.20}$$
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 $\partial^2\phi$?
 A: The Taylor expansion of the superpotential around $\Phi=\phi$ is
$$
W(\Phi)=W(\phi) + 
\left.\frac{\partial W}{\partial \Phi}\right|_{\Phi=\phi}(\Phi - \phi) +
\frac12 \left.\frac{\partial^2 W}{\partial \Phi^2}\right|_{\Phi=\phi}(\Phi - \phi)^2 + \cdots
$$
If you write a chiral field as a function of the co-ordinates $y$ and $\theta$, 
$$
\Phi(y, \theta)  = \phi(y) + \sqrt{2}\theta\psi(y)- \theta\theta F(y)
$$
You find that,
$$
W(\Phi) = W(\phi) + 
\left.\frac{\partial W}{\partial \Phi}\right|_{\Phi=\phi} (\sqrt{2}\theta\psi(y) - \theta\theta F(y)) +
\left.\frac{\partial^2 W}{\partial \Phi^2}\right|_{\Phi=\phi} \theta\theta\psi(y)\psi(y) 
$$
In other words, all the derivatives are wrt the superfield $\Phi$ and not the scalar component $\phi$. Note that the Taylor expansion terminates because of the Grassman co-ordinates.
For what it's worth, $\phi$ and $\partial^\mu \phi$ are independent - this is the question of whether speed and position are independent co-ordinates in classical mechanics.
