Well, it's reassuring for textbook/manuscript authors if the readers take the typos with good humor ;-)). I wish I could deliver typo-free texts though...
One should, however, mention that it's the crucial point of "gauging" a global symmetry to a gauge invariance that the gauge-covariant derivative transforms as the field for spacetime-dependent phase factors, i.e., you introduce the gauge field ##A_{\mu}(x)## as a connection to define the gauge-covariant derivative, i.e.,
$$\mathrm{D}_{\mu}=\partial_{\mu} + \mathrm{i} q A_{\mu}.$$
It's defined such that for
$$\psi'(x)=\exp[-\mathrm{i} q \alpha(x)]$$
the covariant derivative also fulfills the same transformation law,
$$\mathrm{D}_{\mu}' \psi'(x)=\exp[-\mathrm{i} q \alpha(x)] \mathrm{D}_{\mi} \psi(x),$$
and the question is how ##A_{\mu}## transforms to ##A_{\mu}'##. Indeed
$$\mathrm{D}_{\mu}' \psi'(x)=(\partial_{\mu} + \mathrm{i} q A_{\mu}') [\exp(-\mathrm{i} q \alpha) \psi] = \exp(-\mathrm{i} q \alpha) [\partial_{\mu} +\mathrm{i} q (A_{\mu}'-\partial_{\mu} \alpha)] \psi$$
Since this should be
$$\exp[-\mathrm{i} q \alpha(x)] \mathrm{D}_{\mi} \psi(x) = \exp[-\mathrm{i} q \alpha(x)] (\partial_{\mu} +\mathrm{i} q A_{\mu}) \psi(x),$$
you get
$$A_{\mu}'-\partial_{\mu} \alpha=A_{\mu} \; \Rightarrow \; A_{\mu}'=A_{\mu} + \partial_{\mu} \alpha.$$
