Basis for $T_x \mathbb{R}$ Consider the manifold $\mathbb{R}$ with coordinate chart $(\mathbb{R}, \psi)$ where $\psi(x)= x^3$. 
I am looking at the tangent space of $\mathbb{R}$ with respect to the chart $\psi$. 
Lee states that for any $x \in \mathbb{R}$, $\frac{\partial}{\partial x^1}$ will form a basis for $T_x \mathbb{R}$. In this case mentioned above I think $x^1 = x^3$. 
However, I am not sure how to interpret $\frac{\partial}{\partial x^3}$ as a basis for the tangent space. In particular, what does it mean to take the derivative with respect to the coordinate function $x^i$? 
Is there a geometric interpretation of this? 

Remark: If $c:(-\epsilon, \epsilon) \to M$ and $t$ is the parameter (i.e coordinate function for $\mathbb{R}$) then we define;
$$c_*\left(\frac{d}{dt}\Bigr|_{p}\right) = c'(p): p \in (-\epsilon, \epsilon)$$


*

*$\psi: \mathbb{R} \to \mathbb{R}$ 

*Let $x,y$ be the coordinate function of $\mathbb{R}$ in the domain, range respectively

*Then $\dfrac{d}{dx}\Bigr|_p, \dfrac{d}{dy}\Bigr|_{\psi(p)}$ form basis for the tangent space of $\mathbb{R}$ at their respective points

*The differential $\psi_*: T_p \mathbb{R} \to T_{\psi(p)} \mathbb{R}$ and so;
$$\psi_*\left(\frac{d}{dx}\Bigr|_p\right) = \lambda \cdot \frac{d}{dy}\Bigr|_{\psi(p)}$$


*Apply $y$ to both sides:


$$\psi_*\left(\frac{d}{dx}\Bigr|_p\right)(y) = \frac{d}{dx}\Bigr|_p(y \circ \psi) = \lambda$$


*Here by definition $y \circ \psi = \psi$ and so;


$$\frac{d}{dx}\Bigr|_p(\psi) = 3p^2 = \lambda$$


*Thus $\psi_* = 3p^2 \dfrac{d}{dy}\Bigr|_{\psi(p)}$ and by definition $\psi_* = \psi'$

