In Spacetime & Geometry, Carrol immediately posits that, given that you have some tangent space vector (in the coordinate basis representation):
$\frac{d}{d\lambda}=\frac{dx^{\mu}}{d\lambda}\partial _{\mu}$
then the basis vectors $\partial _{\mu}$ will transform according to the chain rule:
$\partial _{\mu '} =\frac{\partial x^{\mu}}{\partial x^{\mu '}} \partial _{\mu}$
But this doesn't seem immediately obvious to me. I thought I would motivate it like this: say we have coordinates that are related by $x^{\mu '}=x^{\mu '}(x^{\mu})$ and for some smooth function $f: M \rightarrow \mathbb{R}$ and some directional derivative $d / d\lambda$ we have (using the chain rule):
\begin{align}\frac{df}{d\lambda}&=\frac{dx^{\mu}}{d\lambda} \frac{\partial f}{\partial x^{\mu}} \\
&=\frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{d x^{\mu '}}{d x^{\lambda}} \frac{\partial f}{\partial x^{\mu '}}\frac{\partial x^{\mu'}}{\partial x^{\mu }}
\end{align}
From which we can identify:
$\partial _{\mu}=\frac{\partial x^{\mu'}}{\partial x^{\mu }} \partial _{\mu '}$
and thus:
$\partial _{\mu '}=\frac{\partial x^{\mu}}{\partial x^{\mu '}} \partial _{\mu }$
Is this a good motivation? Also, it is not clear to me when to use total and partial derivatives here. I felt that anything w.r.t. $\lambda$ should be a total derivative, but I am not sure why.
My first approach was some convolved diagram with multiple coordinate charts on my manifold and then somehow using that to set up some equation but that turned out to be a huge mess.
