I try currently to revise my classic. Could you help me please to demonstrate :
$$\frac{\partial T^{\prime\mu}}{\partial x^{\prime \alpha}}=\frac{\partial}{\partial x^{\prime \alpha}}\left(\frac{\partial x^{\prime \mu}}{\partial x^{v}} T^{v}\right)$$
i.e by using a changing of coordinates between $x^{\mu}$ and $x^{\prime\mu}$. $T^{\mu}$ represents the contravariant components of a "vector" (I mean a tensor of rank 1).
This demostration would be the same than proving :
$$T'^{\mu}\text{d}x^{\nu}=T^{\nu}\text{d}x^{\prime\mu}$$
How to justify this equality if I apply a 1-form $\text{d}x^{\mu}$ or another 1-form $\text{d}x^{\prime\mu}$ on a vector $T$ ?
This appears to be a little over-complicated. You have a basis change with coordinate transformation $(x')^μ=C^μ{}_νx^ν$. Accordingly the contravariant tensor $T$ transforms as $(T')^μ=C^μ{}_νT^ν$. Now compute partial derivatives using the chain rule and $x=C^{-1}x'$ $$ \frac{∂(T')^μ}{∂(x')^α}=\frac{∂(C^μ{}_νT^ν)}{∂x^κ}\frac{∂x^κ}{∂(x')^α}=C^μ{}_ν\,(C^{-1})^κ{}_α\frac{∂T^ν}{∂x^κ} $$
Let's try in a larger context. ${\bf T}=T^ν{\bf e}_ν=T'^μ{\bf e}'_μ$ is a vector field in tangent space. The basis vectors are the coordinate derivatives ${\bf e}_ν=\frac{∂}{∂x^ν}$. In that sense one gets by the chain rule $$ {\bf e}_ν=\frac{∂x'^μ}{∂x^ν}{\bf e}'_μ $$ which then gives $$ T'^μ=T^ν\frac{∂x'^μ}{∂x^ν} $$ Now put parentheses around it and the derivative operator for $x'^α$ before it to get the claimed formula.
If the coordinate change is linear as in the first try, then $\frac{∂x'^μ}{∂x^ν}=C^μ{}_ν$ and $$ C^μ{}_κ\frac{∂x^κ}{∂x'^α}=\frac{∂(C^μ{}_κx^κ)}{∂x'^α} =\frac{∂x'^μ}{∂x'^α}=\delta_α^μ \implies \frac{∂}{∂x'^α} = \frac{∂x^κ}{∂x'^α}\frac{∂}{∂x^κ} =(C^{-1})^κ{}_α\frac{∂}{∂x^κ} $$