# Differential of smooth function on manifold

In the book I am using, the author defines differentials in the following way. Given smooth manifolds $M,N$ and a smooth mapping $\psi:M\to N$ define the differential $d\psi_m$ at a point $m\in M$ as the following linear mapping: ($M_n$ denotes the tangent space at n, defined here as the space of linear derivations on germs of $m$) $$d\psi_m:M_m\to N_{\psi(m)}$$ with $d\psi_m(v)g:=v(g\circ\psi)$ for germs $g$ on $\psi(m)$.

If $N=\Bbb R,\,\,\,\,\,d\psi(v)=v(\psi)\frac{d}{dx}|_{\psi(m)}$ which is "identified" with the dual of $d\psi_m$ evaluated at the dual vector of $\frac{d}{dx}|_{\psi(m)}$, namely $d\psi_m(v)=v(f)$ . Why can we identify them?

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In general, the differential maps each tangent space of $M$ linearly into the appropriate tangent space of $N$. But when $N=\mathbb R$, each tangent space $N_{\psi(m)}$ is canonically isomorphic (as a vector space) to $\mathbb R$, via the map $a\frac{\partial}{\partial x}_{\big| x=\psi(m)} \mapsto a$. When two objects are canonically isomorphic to one another, we can think of them as being the same, as long as we consider only structures covered by the notion of isomorphism at hand. Therefore, we can think of $d\psi_m$ as a linear map from $M_m$ to $\mathbb R$.
On the other hand, we already have a linear map from $M_m$ to $\mathbb R$ — the evaluation of tangent vectors on $\psi$, namely $v\mapsto v(\psi)$. Looking at the definition of $d\psi_m$ again, we realize it's the same map. So... much ado about nothing, really.
Just to build off of this: Ultimately, you're asking why we can identify $v(\psi) \left.\frac{d}{dx}\right.|_{\psi(m)} \in \mathbb{R}_{\psi(m)}$ with $v(\psi) \in \mathbb{R}$. Well, $\mathbb{R}_{\psi(m)}$ is a $1$-dimensional vector space with basis $\{\frac{d}{dx}|_{\psi(m)}\}$ that is canonically isomorphic to $\mathbb{R}$ via the isomorphism described in this answer: $a \left.\frac{d}{dx}\right.|_{\psi(m)} \mapsto a$. – Jesse Madnick May 1 '13 at 21:50