Using the chain rule, show that the derivative transforms as
$$\frac{\partial}{\partial x^\mu}\to\frac{\partial}{\partial x'^\mu}={\Lambda_\mu}^\nu\frac{\partial}{\partial x^\nu}\tag{A}$$
This is the way I would do this, firstly, by the chain rule
$$\frac{\partial}{\partial x'^\mu}=\frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial }{\partial x^\nu}\tag{1}$$
and noting that contravariant vectors transform as $x^\mu \to x'^\mu={\Lambda^\mu}_\nu x^\nu$,
then the inverse Lorentz transformation for contravariant vectors is
$x^\mu={\Lambda_\nu}^\mu x'^\nu$. Relabelling $\mu \to \nu$ and $\nu \to \mu$, then this inverse Lorentz transform can be written as $x^\nu={\Lambda_\mu}^\nu x'^\mu$.
Then $$\frac{\partial x^\nu}{\partial x'^\mu}=\frac{\partial\left({\Lambda_\mu}^\nu x'^\mu\right)}{\partial x'^\mu}\stackrel{\color{red}{?}}={\Lambda_\mu}^\nu\frac{\partial x'^\mu}{\partial x'^\mu}={\Lambda_\mu}^\nu\times 1\tag{2}$$
now by direct substitution into $(1)$, the result, $(\mathrm{A})$ follows immediately. I put a red question mark over the part of my proof for which I am unsure about. Since I am assuming the $4\times 4$ boost matrix, ${\Lambda_\mu}^\nu$ is constant as far as the differentiation is concerned.
Now, the proof given by the author is almost identical to mine, (except for one small change which I don't fully understand the need for):
The chain rule gives $$\frac{\partial}{\partial x'^\mu}=\frac{\partial x^\nu}{\partial x'^\mu}\frac{\partial }{\partial x^\nu}$$
Using $x^\nu={\Lambda_\rho}^\nu x'^\rho$
$$\frac{\partial x^\nu}{\partial x'^\mu}=\frac{\partial}{\partial x'^\mu}{\Lambda_\rho}^\nu x'^\rho\quad{\color{blue}{=}}\quad{\Lambda_\rho}^\nu\delta_\mu^\rho={\Lambda_\mu}^\nu\tag{3}$$
So $$\frac{\partial}{\partial x'^\mu}={\Lambda_\mu}^\nu\frac{\partial}{\partial x^\nu}$$
Where I note that after blue equality in eqn. $(3)$ the author has used the fact that $$\frac{\partial x'^\rho}{\partial x'^\mu}=\delta_\mu^\rho \tag{4}$$
But why did the author need to use $(4)$? This makes me believe that the manipulation I used after the question-marked equality for eqn. $(2)$ is not valid. Put another way, is differentiating wrt. to a set of coordinates, $x'^\mu$, in a new (primed) frame legitimate?
Best Answer
Unfortunately, there are two mistakes in this equation.
First mistake, this expression $$\frac{\partial\left({\Lambda_\mu}^\nu x'^\mu\right)}{\partial x'^\mu}$$ is wrong, because the index $\mu$ appears three times. You should have changed $\mu$ to another letter, say, $\rho$. Thus, you would obtain $$\frac{\partial x^\nu}{\partial x'^\mu}=\frac{\partial\left({\Lambda_\rho}^\nu x'^\rho\right)}{\partial x'^\mu}={\Lambda_\rho}^\nu\frac{\partial x'^\rho}{\partial x'^\mu}$$
Second mistake, this expression $$\frac{\partial x'^\mu}{\partial x'^\mu}=1$$ is wrong. The correct result is $$\frac{\partial x'^\mu}{\partial x'^\mu}=\frac{\partial x'^0}{\partial x'^0}+\frac{\partial x'^1}{\partial x'^1}+\frac{\partial x'^2}{\partial x'^2}+\frac{\partial x'^3}{\partial x'^3}=4,$$ because the convention that repeated indices imply the summation is to be done.
Because $x'^0,x'^1,x'^2,x'^3$ are independent variables. Thus, for example, $$\frac{\partial x'^0}{\partial x'^0}=1 \ ; \frac{\partial x'^0}{\partial x'^1}=0; \frac{\partial x'^0}{\partial x'^2}=0; \frac{\partial x'^0}{\partial x'^3}=0,$$ and so on. Therefore, $$\frac{\partial x'^\rho}{\partial x'^\mu}=\delta^\rho_\mu=\begin{cases}1, \text{if} \ \ \rho=\mu\\0, \text{if} \ \ \rho\ne\mu\end{cases}$$ is correct.