How to calculate.

It is useful for calculations to know how to expand various quantities in these co-ordinate basis. First let $f$ be a smooth function on $M$ then we must have

$$df(x) = \sum_{i=1}^n a_i d\psi^i(x)$$

for some real numbers $a^i$. This is just linear algebra as is the fact that if we apply both sides of this equation to $\partial /\partial \psi^j(x)$ and use the dual basis relation we deduce that

$$a^i = df(x)\left(\frac{\partial\phantom{\psi^i}}{\psi^i}(x)\right)$$

we define

$$\frac{\partial f }{ \partial \psi^i}(x) = df(x)\left(\frac{\partial\phantom{\psi^i}}{\psi^i}(x)\right)$$

and hence have the formula:

$$df(x) = \sum_{i=1}^n \frac{\partial f }{ \partial \psi^i}(x) d\psi^i(x).$$

If $\gamma$ is a path through $x$ then its tangent at 0, $\gamma'(0)$ can be expanded as

$$\gamma'(0) = \sum_{i=1}^n b^i \frac{\partial\phantom{\psi^i}}{\psi^i}(x).$$

Applying $d\psi^j(x)$ to both sides and using the chain rule we deduce that

$$\gamma'(0) = \sum_{i=1}^n (\psi^i\circ\gamma)'(0) \frac{\partial\phantom{\psi^i}}{\psi^i}(x).$$