In a very old Spiegel book I read this very interesting exercise:
Actually $v^k = \frac{dx^k}{dt}$ is a tensor because $\bar{x}^j=\bar{x}^j (x^k)$ so we have $$\bar{v}^j=\frac{d\bar{x}^j}{dt}=\frac{\partial \bar{x}^j}{\partial x^k} \frac{dx^k}{dt} = \frac{\partial \bar{x}^j}{\partial x^k} v^k $$ It is said that $a^k = \frac{\delta \bar{v}^k}{\delta t}$ (in which $\delta$ denotes absolute derivative) too is a tensor. But... why? Trying to obtain acceleration using ordinary derivative (we will use a hat $\hat{\phantom{a}}$ to denote this not so general definition of acceleration) we have (as for the central step see the equation above) $$ \bar{\hat{a}}^j = \frac{d \bar{v}^j}{dt} = \frac{d}{dt} \left( \frac{\partial \bar{x}^j}{\partial x^k} v^k \right) = \frac{\partial \bar{x}^j}{\partial x^k} \hat{a}^k + v^k \frac{d}{dt} \frac{\partial \bar{x}^j}{\partial x^k} $$ Clearly in general $v^k \frac{d}{dt} \frac{\partial \bar{x}^j}{\partial x^k}$ is not zero and so actually this attempt to define an acceleration that transform in the right way failed. But why absolute derivative solve the problem? Saying that $a^k = \frac{\delta \bar{v}^k}{\delta t}$ is a tensor means, in equation, $$ \frac{\delta \bar{v}^p}{\delta t} = \frac{\partial \bar{x}^p}{\partial x^i} \frac{\delta v^i}{\delta t} $$ But how this is the case? I tried to proof that but I can't do it, I did this
Christoffel symbols of second kind cancel out but anyway the final equation doesn't look useful.
Edit
I read in Aris book that acceleration is a tensor:
Who is right? Spiegel or Aris? I'm confused. I can't understand how absolute derivative solve the problem of making double derivative of a parametric curve $x^i=x^i(t)$ with respect its parameter $t$ a tensor. How this works?