Why is the derivative of a tensor not a tensor (in general)? I see that the need to apply the product rule when using curvilinear coordinates results in a term for the derivative of the coefficients that still follows tensor transformation rules, and that it is the derivative of the bases vectors that is the problem. But can I have that product rule spelled out in tensor notation with a plain English explanation of what in the second term breaks the transformation rules for tensors?
For instance what makes the second term below not tensorial specifically?
$$\partial_{\mu'} T_{\nu'}=\frac{ \partial x^\mu}{\partial x^{\mu'}} \frac{ \partial x^\nu}{\partial x^{\nu'}} \partial_\mu T_\nu +\frac{ \partial x^\mu}{\partial x^{\mu'}}T_\nu \partial_\mu\left(\frac{ \partial x^\nu}{\partial x^{\nu'}}\right)  $$
In there the $\partial \mu$ transforms like
$$\partial_{\mu'}=\frac{ \partial x^\mu}{\partial x^{\mu'}}\partial_{\mu}
$$
And the idea is to transform $$\partial_{\mu}T_\nu   $$
Is it the $\mu$ index or both? What summation in that second summand messes things up and why?
 A: Ok, so let's say that $T$ is a one-form as you have there, and let's make $$S_{\mu\nu} = \partial_\mu T_\nu$$explicit. Is it true that the relation $$S_{\mu'\nu'} ={\color{blue}{\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\nu}{\partial x^{\nu'}}S_{\mu\nu}}}$$holds? Absolutely not! Because $$S_{\mu'\nu'} = \partial_{\mu'}T_{\nu'} = \frac{\partial x^\mu}{\partial x^{\mu'}} \partial_\mu\left(\frac{\partial x^\nu}{\partial x^{\nu'}}T_\nu\right) = {\color{blue}{\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^\nu}{\partial x^{\nu'}}S_{\mu\nu}}} + {\color{red}{\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial^2 x^\nu}{\partial x^\mu \partial x^{\nu'}}T_\nu}}. $$
A: The formula
$$\partial_{\mu'} T_{\nu'}=\frac{ \partial x^\mu}{\partial x^{\mu'}} \frac{ \partial x^\nu}{\partial x^{\nu'}} \partial_\mu T_\nu +\frac{ \partial x^\mu}{\partial x^{\mu'}}\,T_\nu\,\partial_\mu\left(\frac{ \partial x^\nu}{\partial x^{\nu'}}\right)  $$
is not tensorial
because of the presence of the second term. If you take instead of the partial derivative $\partial_\mu$ the covariant derivative
$$
\nabla_\mu T_\nu=\partial_\mu T_\nu-\Gamma^\rho_{\mu\nu}T_\rho
$$
you get
$$\tag{1}
\nabla_{\mu'} T_{\nu'}=\frac{ \partial x^\mu}{\partial x^{\mu'}} \frac{ \partial x^\nu}{\partial x^{\nu'}} \nabla_\mu T_\nu
$$
which is how a second rank tensor transforms. The Christoffel symbols that make this work obey
\begin{equation}
\Gamma^{\mu '}_{\alpha '\beta '} = \frac{\partial x^{\mu'}}{\partial x^{\gamma}}\frac{\partial^2x^\gamma}{\partial x^{\beta '}\partial x^{\alpha '}}+\frac{\partial x^{\mu'}}{\partial x^\gamma}\frac{\partial x^{\sigma}}{\partial x^{\alpha '}}\frac{\partial x^p}{\partial x^{\beta '}}\Gamma^{\gamma}_{\sigma p}.
\end{equation}
You can read equation (1) as
$$
\nabla_{\mu'} T_{\nu'}=\frac{ \partial x^\nu}{\partial x^{\nu'}} \nabla_{\mu'} T_\nu\,.
$$
In other words, unlike the partial derivative $\partial_{\mu'}$ the covariant derivative $\nabla_{\mu'}$ can be "pulled through" the term
$$
\frac{ \partial x^\nu}{\partial x^{\nu'}}
$$
as if that was a constant. That's the whole point of introducing the covariant derivative.
