In QED and the basic Higgs mechanism, there is a local gauge transformation where a scalar field $\phi$ is transformed as:
$e^{i\theta\eta(x)} \phi$
The partial derivative of this however makes the above not invariant, and so a covariant derivative is introduced in this way:
$D_\mu e^{i\theta\eta(x)} \phi$=$(\partial_\mu- i \theta A_\mu)$$e^{i\theta\eta(x)} \phi$
So, the derivative remains invariant. However, what if the scalar field is transformed by TWO U(1) symmetries like this:
$e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)} \phi$
This may be a strange symmetry transformation, but I wonder how would one make the derivative of this invariant like this:
$e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)} D_\mu \phi$
For the derivative is now of three different functions which differentiate by the product rule like this:
$f'(x)g(x)h(x)$+$g'(x)f(x)h(x)$+$h'(x)f(x)g(x)$
Thus, the derivative of the function would be:
$i \lambda_1 \eta_1' (x)e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)} \phi$ +$i \lambda_2 \eta_2' (x)e^{i\lambda_2\eta_2(x)} e^{i\lambda_1\eta_1(x)} \phi$+$(\partial_\mu \phi) e^{i\lambda_1\eta_1(x)} e^{i\lambda_2\eta_2(x)}$
So, how would the local gauge invariance derivative be applied in this situation? Would another gauge field be introduced such as $B_\mu$ along with $A_\mu$ ?