Transformation of the derivative of the scalar field in Ramond's book about QFT In the book by Pierre Ramond about quantum field theory, he explores in chapter 1.4 (p.13) the behavior of fields under Poincaré transformations. He starts by explaining that infinitesimal transformations have the following effect on an arbitrary function:
$$f'(x') = f(x) + \delta_0 f(x) + \delta x^\mu \partial_\mu f(x)\tag{1}$$
with $\delta_0 f := f'(x) - f(x)$. The transformation of $x^\mu$ is found to be:
$$\delta x^\mu = \frac{i}{2} \epsilon^{\rho\sigma} L_{\rho\sigma} x^\mu = \epsilon^{\mu\rho} x_\rho \tag{2}$$
Then he shows that, for a scalar field, the spin part of the transformation has to vanish by comparing
$$ \phi'(x')=\phi(x) \iff \delta_0 \phi = - \frac{i}{2} \epsilon^{\rho\sigma} L_{\rho\sigma} \phi \tag{3}$$
with the general form of the Lorentz transformation
$$\delta_0 \text{ (something)} = -\frac{I}{2} \epsilon^{\rho\sigma} M_{\rho\sigma} \text{ (something)} \tag{4}$$
with $M_{\rho\sigma} =  L_{\rho\sigma} + S_{\rho\sigma}$ and $L_{\rho\sigma} = i(x_\rho \partial_\sigma - x_\sigma \partial_\rho)$. So we conclude that the scalar field must have spin-$0$.
Then he goes on showing that the spin part does not vanish for the derivative of a scalar field $\partial_\mu \phi$. There I have a hard time understanding how he shows that. The transformation of $\partial_\mu \phi$ is given by:
$\begin{align} \delta \partial_\mu \phi &= \left[ \delta,\partial_\mu \right]\phi + \partial_\mu \phi \\ &= \left[ \delta x^\nu \partial_\nu , \partial_\mu  \right] \phi \\ &= \epsilon^{\nu\rho} x_\rho \partial_\mu \partial_\nu - \epsilon^\nu_{\ \mu} \partial_\nu \phi \tag{5}\end{align}$
Somehow, in Ramond only the 2nd term survives. Is that because we consider the second derivative as the next order? Or does that somehow vanish? Then he just states that we find:
$$\delta_0 \partial_\mu \phi = -\frac{i}{2} \epsilon^{\rho\sigma} L_{\rho\sigma} \partial_\mu \phi - \frac{i}{2} \left(\epsilon^{\rho\sigma} S_{\rho\sigma} \right)_ \mu^{\ \nu} \partial_\nu \phi \tag{6}$$
with
$$\left(S_{\rho\sigma} \right)_\mu^{\ \nu} = i \left( g_{\rho\mu} g^\nu_{\ \sigma} - g_{\sigma \mu} g^\nu_{\ \rho} \right). \tag{7}$$
I could not manage to show that. So my question would be: how did that happen? What spin does the derivative carry, and why?
 A: So here is the answer to about 3/4 of my own question. First, let me detail the calculation of $\delta \partial_\mu \phi$:
$$\begin{align} \delta \partial_\mu \phi &= \left[\delta,\partial_\mu \right] \phi + \partial_\mu \underbrace{\delta \phi}_{=0} \\ &= \underbrace{\left[ \delta_0,\partial_\mu \right]}_{=0} \phi + \left[\delta x^\nu \partial_\nu , \partial_\mu \right] \phi \\ &= \delta x^\nu \partial_\mu \partial_\nu \phi - \partial_\mu \left( \delta x^\nu \partial_\nu \phi \right) \tag{8} \end{align}$$
Now we can use eq. (1) (see OP) and also obtain:
$$\delta \partial_\mu \phi = \delta_0 \partial_\mu \phi + \delta x^\nu \partial_\nu \partial_\mu \phi \tag{9}$$
Equating eq. (8) and (9) gives us:
$$\delta_0 \partial_\mu \phi = -\partial_\mu \left( \delta x^\nu \partial_\nu \phi \right) \tag{10}$$
Now we want to try to massage this expression so that we can retrieve the expression given in eq. (6) in the OP. This is actually straightforward:
$$\begin{align} \delta_0 \partial_\mu \phi &= -\partial_\mu \left( \delta x^\nu \partial_\nu \phi \right) \\ &= - \partial_\mu \delta x^\nu \partial_\nu \phi - \delta x^\nu \partial_\mu \partial_\nu \phi \\ &= -\frac{i}{2} i \left( \epsilon_\mu^{\ \nu} \partial_\nu - \epsilon^\nu_{\ \mu} \partial_\nu \right) \phi - \frac{i}{2} \epsilon^{\rho\nu} \underbrace{i \left(x_\rho \partial_\nu - x_nu \partial_\rho \right)}_{=L_{\rho\nu}} \partial_\nu \phi \\ &= -\frac{i}{2} \epsilon^{\rho\sigma} \underbrace{i \left( g_{\rho\mu} g_\sigma^{\ \nu} - g_{\sigma\mu} g_\rho^{\ \nu} \right)}_{=\left(S_{\rho\sigma}\right)_\mu^{\ \nu}} \partial_\nu \phi - \frac{i}{2} \epsilon^{\rho\nu} L_{\rho\nu} \partial_\mu \phi \\ &= -\frac{i}{2} \epsilon^{\rho\sigma} \left(S_{\rho\sigma}\right)_\mu^{\ \nu} \partial_\nu \phi - \frac{i}{2} \epsilon^{\rho\nu} L_{\rho\nu} \partial_\mu \phi \tag{11} \end{align}$$
which is exactly what was wanted in the OP. The only thing that I still cannot answer is, how do we know that this is a spin $1$ object (and is that even the case?)?
