How does the partial derivative of a tensor of rank $n$ creates a tensor of rank $n+1$? (cartesian coordinates)

Question

The partial derivative of a tensor of rank $n$, $T_{...i}$, with respect to $x^j$ can be expressed using the transformation rule:

begin{equation}
frac{partial}{partial x^j}T'_{...i}=frac{partial}{partial x^j}sum_{...k}...frac{partial x^k}{partial x'^i}T_{...k}
end{equation}

Since the derivative is linear:

begin{equation}
frac{partial}{partial x^j}T'_{...i}=sumfrac{partial}{partial x^j}(...frac{partial x^k}{partial x'^i}T_{...k}) 
end{equation}

If I'm correct, applying the product rule and knowing that $frac{partial x^i}{partial x^j}=delta^i_j$ yields:

begin{equation}
frac{partial}{partial x^j}T'_{...i}=sum_{...k}...frac{partial x^k}{partial x'^i}frac{partial T_{...k}}{partial x^j}
end{equation}

I'm having trouble interpreting the result. I believe it's not even correct. What's the $frac{partial T_{...k}}{partial x^j}$ factor?

Cryo · Answer

I think the way to proceed is:

$frac{partial}{partial x^j} T'_{i}=frac{partial}{partial x^j} left(frac{partial x^s}{partial x'^i}T_{s}right)=frac{partial}{partial x^j} left(frac{partial x^s}{partial x'^i}right)T_{s}+frac{partial x^s}{partial x'^i}frac{partial T_{s}}{partial x^j}=frac{partial x'^q}{partial x^j} ,frac{partial^2 x^s}{partial x'^i partial x'^q},T_{s}+frac{partial x^s}{partial x'^i}frac{partial T_{s}}{partial x^j}$

I don't think you get cancellation there. Casing point. Polar and Cartesian coordinates in 2d:

$frac{partial}{partial phi},frac{partial r}{partial x}=frac{partial}{partial phi},cosphi=-sinphineq0$

How does the partial derivative of a tensor of rank $n$ creates a tensor of rank $n+1$? (cartesian coordinates)

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