Prove derivative of contravariant tensor of rank 1 is a mixed tensor of rank 2 $A^\alpha$ is a given contravariant vector (when $\alpha\in {0,1,2,3}$ a $4$-vector in Minkowski space) 
I need to show that the derivative $\frac{\partial A^\alpha}{\partial x^\beta}$ is a mixed tensor of rank 2, so:
$\frac{\partial A^\alpha}{\partial x^\beta}=T^\gamma_\delta$
(I'm not sure if $\gamma=\alpha$ and $\delta=\beta$)
The subject is Special Relativity so the space taking diffrential 2-form.
 A: The most straight-forward, brute-force way to show whether something is a tensor is to perform a change of coordinates and see how the components change.
Given concrete coordinates $x^\beta$ and an arbitrary point $p$ in our space, $A^\alpha$ is a $4$-tuple of numbers (some might call it a vector, or even a column vector), and $T^\alpha_\beta$ is a $4\times 4$ table of numbers describing how the $4$-tuple of $A^\alpha$ change when we move from $p$ to a point with slightly changed $i$th coordinate.
Now, let's say we have a new set of coordinates, $x^{\beta'}$. Since $A$ is a tensor, the $4$-tuple $A^{\alpha'}$ of numbers it gives at point $p$ with the new coordinates is given by
$$
A^{\alpha'} = A^\alpha\frac{\partial x^{\alpha'}}{\partial x^\alpha}
$$
(I have barely touched tensors in two years, and I still remember which coordinate functions to differentiate with respect to which. Such is the power of index notation and knowing how indices must match up.)
If we now differentiate both sides with respect to the new coordinate systems, we get
$$
T^{\alpha'}_{\beta'} = \frac{\partial A^{\alpha'}}{\partial x^{\beta'}}
 = \frac{\partial}{\partial x^{\beta'}} \left(A^\alpha\frac{\partial x^{\alpha'}}{\partial x^\alpha}\right)\\
=  \frac{\partial A^{\alpha}}{\partial x^{\beta'}} \frac{\partial x^{\alpha'}}{\partial x^\alpha} + A^{\alpha}\frac{\partial}{\partial x^{\beta'}}\frac{\partial x^{\alpha'}}{\partial x^\alpha}
$$
By the chain rule, the first term here is
$$
\frac{\partial A^{\alpha}}{\partial x^{\beta'}} \frac{\partial x^{\alpha'}}{\partial x^\alpha} 
= \frac{\partial A^{\alpha}}{\partial x^{\beta}}\frac{\partial x^{\beta}}{\partial x^{\beta'}} \frac{\partial x^{\alpha'}}{\partial x^\alpha}
= T^\alpha_\beta\frac{\partial x^{\beta}}{\partial x^{\beta'}} \frac{\partial x^{\alpha'}}{\partial x^\alpha}
$$
which is the correct change-of-coordinates expression for a $(1, 1)$ tensor like $T$. However, there is a second term:
$$
A^{\alpha}\frac{\partial}{\partial x^{\beta'}}\frac{\partial x^{\alpha'}}{\partial x^\alpha}
$$
And we sadly cannot expect this to be non-zero in general. We can see this, for instance, in $2$ dimensions with Cartesian versus polar coordinates, where
$$
\frac{\partial}{\partial\theta}\frac{\partial r}{\partial x}
$$
clearly cannot be $0$, as $\frac{\partial r}{\partial x}$ is $0$ for a point on the $y$ axis and $1$ for a point on the (positive) $x$-axis, so as a function of $\theta$ it cannot be constant.
Which is to say $T^\alpha_\beta$ is not a tensor. However, there is a way to mend this. The mend comes in the form of the so-called Christoffel symbols and the covariant derivative. Very briefly, the covariant derivative consists of taking the partial derivative, then for each index throw in a correction term using the Christoffel symbols (which have three indices: $\Gamma^\gamma_{\mu\nu}$, and is thus a $4\times 4\times 4$ table of numbers at each point) to ensure that what you get out in the end is indeed a tensor. The covariant derivative of $A^\alpha$ is
$$
\nabla_\beta A^\alpha = \frac{\partial A^\alpha}{\partial x^\beta} + \Gamma^\alpha_{\beta\eta}A^\eta
$$
