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

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

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

Since the derivative is linear:

$$$$\frac{\partial}{\partial x^j}T'_{...i}=\sum\frac{\partial}{\partial x^j}(...\frac{\partial x^k}{\partial x'^i}T_{...k})$$$$

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

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

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?

• You mean the covariant derivative.
– J.G.
Jul 25 at 6:28

$$\frac{\partial}{\partial x^j} T'_{i}=\frac{\partial}{\partial x^j} \left(\frac{\partial x^s}{\partial x'^i}T_{s}\right)=\frac{\partial}{\partial x^j} \left(\frac{\partial x^s}{\partial x'^i}\right)T_{s}+\frac{\partial x^s}{\partial x'^i}\frac{\partial T_{s}}{\partial x^j}=\frac{\partial x'^q}{\partial x^j} \,\frac{\partial^2 x^s}{\partial x'^i \partial x'^q}\,T_{s}+\frac{\partial x^s}{\partial x'^i}\frac{\partial T_{s}}{\partial x^j}$$
$$\frac{\partial}{\partial \phi}\,\frac{\partial r}{\partial x}=\frac{\partial}{\partial \phi}\,\cos\phi=-\sin\phi\neq0$$