Differential of a $C^\infty$ function from a manifold $M$ to $\mathbb{R}$

Question

I am studying differentiable manifolds from Warner. In the book, the differential is defined as follows.

If $\psi:M\longrightarrow N$ is $C^\infty$, and if $m\in M$, then the differential of $\psi$ at $m$ is the linear map $d\psi:M_m\longrightarrow N_{\psi(m)}$, defined as under : If $v\in M_m$, then, we define $d\psi(v)(g) = v(g\circ\psi)$ where $g$ is a $C^\infty$ function on a neighbourhood of $\psi(m)$.

My question is this. They have mentioned this special case : If $f:M\longrightarrow\mathbb{R}$ is a $C^\infty$ function, then $df(v)=v(f)\frac{d}{dr}|_{f(m)}$. I don't know how to show this. By definition, $df(v)(g)=v(g\circ f)$, where $g$ is differentiable in a neighbourhood of $f(m)$. How is $v(g\circ f) = v(f)\frac{dg}{dr}|_{f(m)}$. [I understand the fact intuitively too, because $df(v)$ is a tangent vector on $\mathbb{R}$, and this is like the chain rule, but how does one show it?]

Answer 1

If $U\subset \mathbb R^n$ is open and $f:U\to \mathbb R,\; g:\mathbb R \to \mathbb R$ are differentiable functions we have for $u_0\in U$ : $$\frac {\partial (g\circ f)(u_0)}{\partial x_i}=g'(r_0)\cdot \frac {\partial f(u_0)}{\partial x_i}$$ where $r_0=f(u_0)$

Your result follows from there if you write $$ v=\sum v_i \frac {\partial }{\partial x_i} |_{m_0} \quad (\star) $$ for a tangent vector $v\in T_{m_0}M$ written out in terms of a chart $(x_1,...,x_n)$ for $M$ at $m_0$ and apply both sides of $(\star)$ to $g\circ f$.