In the context of general relativity (GR) it is necessary to introduce the notion of covariant derivatives. From the point of view of a basic introduction, we always start to deal with GR in a highly chart-dependent exposition.
Now, in order to derive the covariant derivative we can simply ask ourselves what is the derivative of a vector field $V$. In other words, we derive the covariant derivative as:
$$\frac{\partial V}{\partial x^{\mu}} = \frac{\partial }{\partial x^{\mu}}\Big(V^{\nu}\frac{\partial }{\partial x^{\nu}}\Big) = \frac{\partial V^{\nu} }{\partial x^{\mu}}\frac{\partial }{\partial x^{\nu}}+ V^{\nu}\frac{\partial }{\partial x^{\mu}}\Big(\frac{\partial }{\partial x^{\nu}}\Big) \implies $$
$$\frac{\partial V}{\partial x^{\mu}} = \Bigg(\frac{\partial V^{\nu} }{\partial x^{\mu}}+ \Gamma^{\nu}\hspace{0.1mm}_{\mu\alpha} V^{\alpha}\Bigg)\frac{\partial }{\partial x^{\nu}} := \nabla_{\mu}V^{\nu}\frac{\partial }{\partial x^{\nu}} \tag{1}$$ Therefore we have derived the covariant derivative:
$$\nabla_{\mu}V^{\nu} = \frac{\partial V^{\nu} }{\partial x^{\mu}}+ \Gamma^{\nu}\hspace{0.1mm}_{\mu\alpha} V^{\alpha} \tag{2}$$
I would like to know: how can I derive the (gauge) covariant derivative given by $(3)$ using more mathematical arguments than physical ones? :
$$D_{\mu} = \partial_{\mu} - iqA_{\mu} \tag{3}$$
