Peskin and Schroeder Chapter 19 anomalies 19.63 Lagrangian

Question

I am (self) studying chapter 19 of Peskin and Schroeder's Introduction to Quantum Field Theory. Around equation (19.63) they state the Lagrangian is invariant if $\alpha$ is a constant, and if $\alpha(x)$ is dependent on $x$ they obtain the top line of (19.63). I am not sure how to obtain the $\partial_{\mu}\alpha(x)$ term. I have attempted to substitute in the transformations (19.62) into the Lagrangian, but I have ended up with $i\bar{\Psi}\alpha(x)\not \!\!D \alpha(x)\Psi$ when I expand out these brackets. The transformations in equations (19.62) are $\Psi(x)\rightarrow\Psi(x)'=(1+i\alpha(x)\gamma^{5})\Psi(x), \bar{\Psi}\rightarrow\bar{\Psi}(x)'=\bar{\Psi}(1+i\alpha(x)\gamma^{5})$. They get the Lagrangian transforms to $i\bar{\Psi}\not{\!\!D}\Psi-\partial_{\mu}\alpha(x)\bar{\Psi}\gamma^{\mu}\gamma^{5}\Psi$

My calculation is the following for the Lagrangian density: $L'=i\bar{\Psi}(1+i\alpha\gamma^{5})\gamma^{\mu}D_{\mu}(1+i\alpha\gamma^{5})\Psi=i\bar{\Psi}(1+i\alpha\gamma^{5})(1-i\alpha\gamma^{5})\gamma^{\mu}D_{\mu}\Psi$ using that $\gamma^{\mu}$ and $\gamma^{5}$ anti-commute. Then I expanded the brackets and used that $(\gamma^{5})^{2}=1$ $L'=i\bar{\Psi}(1+i\alpha\gamma^{5}-i\alpha\gamma^{5}+\alpha^{2}(\gamma^{5})^{2})\not{\!\!D}\Psi=i\bar{\Psi}(1+\alpha^{2})\not{\!\!D}\Psi$ is what I get for $\alpha$ constant, which does not give an invariant Lagrangian which is what I need, We neglect $\alpha^{2}$ as small for infinitesimal transformations(Edit: Cosmas Zachos)

For $\alpha(x)$ dependent on $x$: I use a similar method:

$L'=i\bar{\Psi}(1+i\alpha(x)\gamma^{5})\gamma^{\mu}D_{\mu}(1+i\alpha(x)\gamma^{5})\Psi=i\bar{\Psi}(1+i\alpha(x)\gamma^{5})(1-i\alpha(x)\gamma^{5})\gamma^{\mu}D_{\mu}\Psi$ using that $\gamma^{\mu}$ and $\gamma^{5}$ anti-commute. Then I expanded the brackets and used that $(\gamma^{5})^{2}=1$

$L'=i\bar{\Psi}(1+i\alpha(x)\gamma^{5}-i\alpha(x)\gamma^{5}+\alpha^{2}(x)(\gamma^{5})^{2})\not{\!\!D}\Psi=i\bar{\Psi}(1+\alpha^{2}(x))\not{\!\!D}\Psi$ $L'=i(\bar{\Psi}\gamma^{\mu}D_{\mu}\Psi+i\bar{\Psi}\alpha(x)\gamma^{5}\gamma^{\mu}D_{\mu}\Psi+i\bar{\Psi}\gamma^{\mu}D_{\mu}\alpha(x)\gamma^{5}\Psi-\bar{\Psi}\alpha(x)\gamma^{5}\gamma^{\mu}D_{\mu}\alpha(x)\gamma^{5}\Psi$ then using anti-commutation relations again gives my result $L'=i\bar{\Psi}\gamma^{\mu}D_{\mu}\Psi+i\bar{\Psi}\alpha(x)\not \!\! D\alpha(x)\Psi$

Answer 1

$$i\bar{\Psi}(1+i\alpha(x)\gamma^{5})\not{\!\!D}\left ((1+i\alpha(x)\gamma^{5})\Psi\right)\\ =i\bar{\Psi} \not{\!\!D} \Psi -\bar{\Psi} (\not\!\partial \alpha)\gamma^5 \Psi +O(\alpha^2) \\= i\bar{\Psi} \not{\!\!D} \Psi -\partial_{\mu}\alpha(x)\bar{\Psi}\gamma^{\mu}\gamma^{5}\Psi +O(\alpha^2),$$ where you have used Leibniz's rule of differentiation for $\partial (\alpha \Psi)$ and discarded the $O(\alpha^2)$ as you are meant to, for infinitesimal transformations.