Partial derivatives on manifolds in terms of local charts

Question

Let $\phi=(u^1,\cdots, u^n)$ be a coordinate system in manifold $M$ at $p$. If $f \in c^{\infty}(M)$, we define $$\frac{\partial f}{\partial u^i} (p) = \frac{\partial(f \circ \phi ^{-1})}{\partial x^i} \qquad 1 \le i \le n.$$ Where $x^1, \cdots , x^n$ are the natural coordinate functions of $R^n$. Could we rewrite that like the following: $$\frac{\partial f}{\partial u^i} (p) = \frac{d(f \circ (u^i) ^{-1})}{d t} \qquad 1 \le i \le n.$$

Answer 1

No. $u^i$ is the $i$th coordinate function of $\phi$, and as such it's a $C^\infty$ map from an open subset of $M$ into $\mathbb R$. Thus it cannot be bijective and so it cannot have an inverse function.