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.
Best Answer
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.