What's the derivative of a map defined on manifolds?

Question

I'm going through Warner's book on differentiable manifolds. On page 8 he defines what it means for a map $f: U \subset M \to \mathbb R$ to be differentiable:

$f$ is differentiable iff $f \circ \psi$ is differentiable for all charts $\psi$ on $M$.

He does not proceed to give a definition of the derivative of $f$. I tried to do a web search but did not find a definition. Is the derivative of $f$ just defined to be the derivative of $f \circ \psi$?

What's the definition of the derivative of a map defined on manifolds?

Edit

At the bottom of page 105 in this book (in the proof of the regular level set theorem) the author calculates the Jacobian of a map $F: M \to N$. This Jacobian contains entries of the form ${\partial F \over \partial x_i}$. So, it seems to me that the derivative, at least partial derivatives exist (although a comment below by Mariano Suarez-Alvarez suggests otherwise)

Answer 1

Let $f:M\rightarrow \mathbb{R}$ be a function from a manifold M to the manifold $\mathbb{R}$. Also, let $\gamma$ be a smooth curve on the manifold $M$ with $\dot{\gamma}\in TM$ it's derivative.

Then, we define the derivative map of $f$ as the function that does the following \begin{equation} \begin{aligned} Df : TM & \rightarrow T\mathbb{R},\\ \dot{\gamma} & \mapsto df(\dot{\gamma})=\dot{\gamma}(f)= \dot{(f(\gamma))} \end{aligned} \end{equation}

Answer 2

This is essentially the same answer as @ser969547's.

Given a point $p \in M$ and a tangent vector $v \in T_pM$, the directional derivative of a smooth function $f: M \rightarrow \mathbb{R}$ at $p$ in the direction $v$ is defined to be $$ D_vf(p) = \left.\frac{d}{dt}\right|_{t=0}\gamma(t)m $$ where $\gamma$ is a smooth curve such that $\gamma(0) = p$ and $\gamma'(0) = v$. This definition is independent of coordinates. You can check that, with respective to any set of local coordinates, it is the standard definition of a directional derivative for a function on $\mathbb{R}^n$. Moreover, using coordinates you can verify that if you fix $p$, then $$ v \mapsto D_vf(p) $$ is a linear function of $v$. This map is called the differential of $f$ at $p$. It is the appropriate definition of the derivative of $f$. If you write this out in local coordinates, you get the familiar formula that, if $f$ is a function on $\mathbb{R}^n$, then $$ D_vf(p) = v\cdot\nabla f(p). $$ Therefore, $df$ is the coordinate-free version of the gradient.