Determining Feynman rules from Lagrangian To preface, I'm not a physics major so I've never taken a QFT class before. But I'm working on something that requires me to find the amplitude of that following Lagrangian:
$$ \mathcal{L} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} +\bar\psi(i\partial_s -eA_s - m)\psi + \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m_\phi \phi^2 - \frac{i\lambda}{\sqrt{2}}\phi\bar\psi\gamma^5 \psi$$
where $\partial_s$ and $A_s$ is the Feynman slash notation, $\psi$ is a fermion of charge $e$ and mass $m$, $\phi$ is a psuedoscalar, and the last term is a Yukawa potential.
From what I've read online, it seems that to find the amplitude, I need to first determine the Feynman rules from this Lagrangian. Then use those rules to determine the amplitude.
Since this is a QED Lagrangian + a Yukawa potential, can I find the Feynman rules for the QED Lagrangian and the rules for the Yukawa potential separately and then somehow combine them? If not, how should I begin to determine the Feynman rules for this Lagrangian?
 A: The Lagrangian consists of quantum electrodynamics and free scalar field theory with an additional interaction, so you can use the QED Feynman rules. The scalar field has propagator,
$$\frac{i}{p^2-m_\phi^2}.$$
As for the interaction term, this corresponds to a vertex involving a $\bar\psi$, $\psi$ and $\phi$, i.e. an electron, anti electron and scalar particle, to which we associate the factor,
$$\frac{\lambda}{\sqrt 2} \gamma^5.$$
In terms of mass dimension, $[\phi] = [\psi] = [\bar\psi] = 1$, and so the Yukawa interaction requires a coupling constant with $[\lambda] = 1$ which corresponds to a super renormalizable interaction, though the theory as a whole is renormalizable. 
A: You are correct, you can combine the Feynman rules. In the definition of the propagator you only use the free Lagrangian, and then treat each vertex separately. Obviuosly, you are opening to new interactions, and you are also spoiling renormalizability of the theory, but if you only need to do basic calculations there is no need for worry.
