Question about the derivative of a contravariant tensor

Question

I need to show that the derivative of a contravariant tensor ($\frac{\partial T^i}{\partial x^j}$) is not a tensor. With $J^{i'}_i=\frac{\partial x^{i'}}{\partial x^i}$ and $J^{i'}_{ij}=\frac{\partial^2 x^{i'}}{\partial x^i\partial x^j}$, I proceeded as follows

$\dfrac{\partial T^{i'}}{\partial x^{j'}} =\dfrac{\partial}{\partial x^{j'}} \left ( T^i J^{i'}_i \right) = \dfrac{\partial T^i}{\partial x^j} J^{j}_{j'} J^{i'}_i + T^i J^{i'}_{ij}$

Which shows that $\frac{\partial T^i}{\partial x^j}$ is not a tensor. But the right answer is

$\dfrac{\partial T^{i'}}{\partial x^{j'}}=\dfrac{\partial T^i}{\partial x^j} J^{j'}_{j} J^{i'}_i + T^i J^{i'}_{ij}$ (where the first term here is different from my answer).

What is wrong in my calculation?

Answer 1

Both of the answers you suggested are wrong, and you can easily see that by verifying that the indices of the left and right sides of the equation do not match.

There is two types of tensor, covariant and contravariant, and they are classified by the way they transform,

the contravariant tensor transformation : $$ T^{i'}=\frac{\partial x^{i'}}{\partial x^i}T^i $$ the covariant tensor transformation : $$ T_{i'}=\frac{\partial x^i}{\partial x^{i'}}T_i $$

for mixed tensors (general case):

$$T^{\alpha' ...\beta'}_{\ \ \ \ \ \ \ \ \ \ i'...j'}=\frac{\partial x^{\alpha'}}{\partial{x^{\alpha}}}\frac{\partial x^{\beta'}}{\partial{x^{\beta}}}...\frac{\partial x^{i}}{\partial{x^{i'}}}\frac{\partial x^{j}}{\partial{x^{k'}}}T^{\alpha ...\beta}_{\ \ \ \ \ \ \ \ \ \ i...j}$$

the proof that the derivative of a vector is not a tensor is :

$$ \frac {\partial T^{i'}}{\partial x^{j'}}=\frac{\partial}{\partial x^j}(T^{i'})\frac{\partial x^j}{\partial x^{j'}}=\frac{\partial}{\partial x^j}(T^{i} \frac {\partial x^{i'}}{\partial x^{i}})\frac{\partial x^j}{\partial x^{j'}}= \frac{\partial T^{i}}{\partial x^j}\boxed{ \frac {\partial x^{i'}}{\partial x^{i}}\frac{\partial x^j}{\partial x^{j'}}}+ T^{i} \frac{\partial}{\partial x^j}(\frac {\partial x^{i'}}{\partial x^{i}})\frac{\partial x^j}{\partial x^{j'}}$$

notice that if the answer was only right-handside term, then you can deduce that it is a tensor. The appearance of second member rejects that.