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In page 488 of Peskin and Schroeder, it is stated (emphasis mine):

It is not difficult to check using (15.27) and (15.21) that, even for finite transformations, the covariant derivative has the same transformation law as the field on which it acts.

I was trying to indeed verify that.

This is what I have tried:

  1. $V\left(x\right)\in SU\left(2\right)^{\mathbb{R}^4}$
  2. $\psi\left(x\right)\mapsto V\left(x\right)\psi\left(x\right)$.
  3. $D_\mu\left(x\right)\equiv\partial_\mu-igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}$.
  4. $igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\mapsto V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\left[V\left(x\right)^\dagger\right]-V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)^\dagger\right]\right\}$

Thus:

\begin{align} D_\mu\left(x\right)\psi\left(x\right) \mapsto & \left\{\partial_\mu -V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\left[V\left(x\right)^\dagger\right]+V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)^\dagger\right]\right\}\right\}V\left(x\right)\psi\left(x\right) = \\\ &= \left[\partial_\mu V\left(x\right)\right]\psi\left(x\right)+V\left(x\right)\partial_\mu\psi\left(x\right)-V\left(x\right)igA_{\mu,\,j}\left(x\right)\frac{\sigma^j}{2}\psi\left(x\right)+V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\psi\left(x\right) = & \\\ &= V\left(x\right)D_\mu\left(x\right)\psi\left(x\right)+\left\{\left[\partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\right\}\psi\left(x\right) \end{align}

So as far as I understand, $\boxed{\left[\partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\stackrel{?}{=}0}$ should be zero.

To prove that I have used the fact that $VV\dagger=1$:

\begin{align} \partial_\mu\left[ V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \partial_\mu\left[ V\left(x\right)1\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \partial_\mu\left[ V\left(x\right)V\left(x\right)^\dagger V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\ \left[\partial_\mu V\left(x\right)\right]V\left(x\right)^\dagger V\left(x\right) +V\left(x\right)\left[ \partial_\mu V\left(x\right)^\dagger \right]V\left(x\right) +V\left(x\right)V\left(x\right)^\dagger\left[ \partial_\mu V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right) &= \\2\left\{\partial_\mu\left[ V\left(x\right)\right] + V\left(x\right)\left\{\partial_\mu\left[V\left(x\right)\dagger\right]\right\}V\left(x\right)\right\}\end{align}

Is this all correct?

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I found this question while struggling with the same issue. The solution turns out to be quite simple. This works for any general gauge group $G$, with elements $g(x),\ g^{-1}(x),$ and $e$. $$ 0 = \partial_\mu(e) = \partial_\mu (gg^{-1}) = (\partial_\mu g) g^{-1} + g \partial_\mu g^{-1} $$ so $$ \partial_\mu g = - g (\partial_\mu g^{-1}) g $$ In your case $g=V$ and $g^{-1}=V^\dagger$, so this gives you the result you're looking for.

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