Why is partial derivative of tensor not a tensor? Partial differentiation of the transformation law
$$
\bar{T}_i = T_r\frac{\partial x^r}{\partial\bar{x}^i}
$$
of a covariant vector yields
$$
\frac{\partial\bar{T}_i}{\partial \bar{x}^k} = \frac{\partial{T_r}}{\partial x^s} \frac{\partial x^r}{\partial{\bar{x}}^i} \frac{\partial{x}^s}{\partial\bar{x}^k} + T_r \frac{\partial^2 x^r}{\partial{\bar{x}^k}\partial{\bar{x}^i}}.
$$
Because of the second term on the right the partial derivative is not a tensor.
Why is this intuitively expected? I thought a tensor was something intrinsic independent of coordinates and thus invariant under coordinate changes. It is not clear to me why a partial derivative fails this.
 A: Yes, the tensor itself is independent of the coordinate system, but the operation of taking a partial derivative is highly dependent on what coordinate system you're using: you vary one of the coordinates while keeping all the other coordinates (in that coordinate system) constant. And this dependence turns out not to be “tensorial”, when you check what happens if you express the derivative in another coordinate system, as you did in the question.
A: The derivative is independent of coordinates if the basis elements are constant (you’re in Euclidean space), but on a general manifold, you have to use the product rule, but the derivative of a tensor only differentiates the components and thus ignores half of the answer.
A: The second term is due to the fact that in a general coordinate system the components of a vector change under a parallel transport. The covariant derivative of a tensor field is what you get when you subtract the contribution due to the parallel transport from the ordinary derivative.
