Show that differential map $df : TM \to \mathbb{R}$ of a smooth map $f: M \to \mathbb{R}$ is smooth. I'm trying to prove that for any smooth map $f: M \to \mathbb{R}$ on a manifold $M$ to the real numbers, the map $df : TM \to \mathbb{R}$ from the tangent bundle $TM$ to the real numbers given by $df(v) = v(f)$ is smooth.
I was told that I need to show that every component of $df$ is smooth, using a coordinate map of the tangent bundle $TM$ (with respect to a chart on $M$ and the natural projection map from $TM$ to $M$).
I'm pretty sure that I need to show that $df$ is smooth with respect to any given smooth vector field on $M$. My plan was to show that the composition of the inverse of a chart of $TM$ and $df$ is a smooth diffeomorphism, as one proves that any vector field can be written as a linear combination of smooth functions and the partial derivitive of the components of a chart. But I'm having a hard time trying to write out each components explicitly. I would love some helps!!
P.S. I'm new here, and I apologive that I don't know how to write mathematical symbols as in LaTeX when asking a question. 
 A: It suffices to check that ${\rm d}f\colon TM \to \Bbb R$ is smooth in enough charts to cover $TM$. If $(U, \varphi)$ is a chart in $M$, we have a induced chart in $TM$ given by $(TU,\widetilde{\varphi})$, given as follows: if $\varphi = (x^1,\ldots, x^n)$, we write a tangent vector as $$v_p = \sum_{i=1}^n a^i \frac{\partial}{\partial x^i}\bigg|_p,$$ and put $\widetilde{\varphi}(v_p) = (x^1(p),\ldots, x^n(p), a^1,\ldots, a^n)$. We of course have that the inverse is $$\widetilde{\varphi}^{-1}(p^1,\ldots, p^n, v^1,\ldots, v^n) = \sum_{i=1}^n v^i \frac{\partial}{\partial x^i}\bigg|_{\varphi^{-1}(p^1,\ldots, p^n )}.$$Then $$\begin{align} {\rm d}f \circ \widetilde{\varphi}^{-1}(p^1,\ldots, p^n, v^1,\ldots, v^n) &= {\rm d}f\left(\sum_{i=1}^n v^i \frac{\partial}{\partial x^i}\bigg|_{\varphi^{-1}(p^1,\ldots, p^n )} \right)\\ &=\sum_{i=1}^n v^i \,{\rm d}f\left(\frac{\partial}{\partial x^i}\bigg|_{\varphi^{-1}(p^1,\ldots, p^n)}\right) \\ &= \sum_{i=1}^n v^i \frac{\partial(f\circ \varphi^{-1})}{\partial x^i}(p^1,\ldots, p^n)\end{align}$$is smooth in the variables $(p^1,\ldots,p^n, v^1,\ldots, v^n)$, since $f$ being smooth tells us that $$(p^1,\ldots, p^n) \mapsto \frac{\partial(f\circ \varphi^{-1})}{\partial x^i}(p^1,\ldots, p^n) $$is smooth.
