Show $df(\frac{\partial}{\partial x_j}\Big|_m ) = \sum_i \frac{\partial(y_i\circ f)}{\partial x_j} \frac{\partial}{\partial y_i}\Big|_{f(m)}$

Question

Given a smooth function $f:(U,\phi = (x_1,\cdots,x_n))\rightarrow (V,\psi = (y_1,\cdots, y_n))$, I would like to show the following remark in Warner's differential manifolds book, $$df(\frac{\partial}{\partial x_j}\Big|_m ) = \sum_i \frac{\partial(y_i\circ f)}{\partial x_j} \frac{\partial}{\partial y_i}\Big|_{f(m)}$$ where $\frac{\partial}{\partial x_j} \in T_m U$ and $\{\frac{\partial}{\partial y_i}\}$ is a basis of $T_{f(m)} V$.

By definition we have $df(v)g = v(g\circ f)$, so now let $g:V\rightarrow \mathbb{R}$, we look at how the left hand side acts on $g$. We have $\frac{\partial}{\partial x_j} (g\circ f)$, and I tried to write it in the standard $\mathbb{R}^n$ coordinate $$\frac{\partial(g \circ f \circ \phi^{-1})}{\partial r_j} $$ but I couldn't get any further.

Edit: a few pages before this, I see that we have the following fact
$$v= \sum_i v(x_i) \frac{\partial}{\partial x_i}$$ which trivialized the remark above. (Here $v(x_i)$ is the tangent vector $v$ acts on $x_i$) However, this proof is complicated and it used the fact that we define the tangent vector as a derivation.

Is there a simpler approach to this problem?

Answer 1

By realizing that $df$ (the notation is bad) is the Jacobian, you should expect to have (I dropped the point and the coordinate maps, which are not essential): $$ df(\frac{\partial}{\partial x_j})=\sum_i\alpha_i\frac{\partial }{\partial y_i} $$ It remains to determine $\alpha_i$. Plug in on both sides the coordinate function $y_i$, and you have RHS$=\alpha_i$. Now the LHS equals: $$ df(\frac{\partial}{\partial x_j})y_i=\frac{\partial}{\partial x_j}(y_i\circ f) =\frac{\partial (y_i\circ f)}{\partial x_j} $$ Can you take it from here?