The derivative of a function.

Recall that a function $f \colon M\to \mathbb{R}$ is smooth if we can cover $M$ with co-ordinates $(U, \psi)$ such that $f \colon \psi^{-1} \colon \psi(M) \to \mathbb{R}$ is smooth . If $\gamma$ is a smooth path through $x \in M$ then if follows from the chain rule that

$$f\circ \gamma=(f\circ\psi^{-1}) \circ (\psi\circ \gamma)$$

is smooth. Hence we can differentiate the function $f \circ \gamma$ at $t=0$. By the chain rule we have that

$$(f\circ\gamma)'(0) = d(f \circ \psi^{-1})(\psi(x))((\psi^i \circ \gamma)'(0)).$$

It follows that $(f\circ\gamma)'(0) = (f\circ\rho)'(0)$ if $\rho$ and $\gamma$ are in the same tangency class. Hence if $X = t_0(X)$ is a tangent vector in $T_xM$ we can define

$$df(x)(X) = (f\circ\gamma)'(0).$$

We call this the rate of change of $f$ in the direction $X$. Notice that we can calculate $df(x)(X)$ without explicit reference to the path $\gamma$ by the formula

$$df(x)(X) = d(f\circ\psi^{-1})(\psi(x)) d\psi(x)(X).$$

As we vary the tangent vector $X$ we define a map

$$df(x) \colon T_xM \to \mathbb{R}$$

called the differential of $f$ at $x$. This map satisfies the formula

$$df(x) = d(f\circ\psi^{-1})(\psi(x)) \circ d\psi(x)$$

and hence, being a composition of linear maps, is linear.

We call the set of linear maps from $T_xM$ to $\mathbb{R}$ the cotangent space to $M$ at $x$ and denote it by $T_x^*M$. So we have

$$df(x) \in T_x^*M.$$

Elements of $T_x^*M$ are also called one-forms.