# Curves and smooth functions on manifolds derivative

Let $$f: M\rightarrow \mathbb{R}$$ be a smooth map at $$p\in M$$ and let $$\gamma:(-\epsilon,\epsilon)\rightarrow M$$ be a smooth curve with $$\gamma(0)=p$$. Let $$(U,\phi)$$ be a chart at $$p$$. Set, $$(\phi \circ \gamma)(t)=(x_1(t),......,x_n(t))$$

We claim:

$$\frac{d}{dt}|_{t=0}(f\circ \gamma)=\sum_i\frac{\partial{f\circ \phi^{-1}}}{\partial{x}_i}|_{\phi(p)}\frac{d{x_i}}{dt}|_{t=0}$$

Is this because,

$$\frac{d}{dt}|_{t=0}(f\circ \gamma)=\frac{d}{dt}|_{t=0}((f\circ \phi^{-1})\circ (\phi \circ \gamma)).$$ and then the chain rule in euclidean space implies what we claim?

Yes. For some $$\delta \in (0,\epsilon)$$ the map $$f\circ \gamma : (-\delta,\delta) \to \mathbb R$$ is the composition of $$\phi \circ \gamma : (-\delta,\delta) \to V \subset \mathbb R^n$$ and $$f\circ \phi^{-1} : V \to \mathbb R$$. The chain rule gives us therefore
$$(f\circ \gamma)'(0) = (f\circ \phi^{-1})'(\phi \circ \gamma(0)) \cdot (\phi \circ \gamma)'(0) = (f\circ \phi^{-1})'(\phi(p)) \cdot (\phi \circ \gamma)'(0)$$ where $$(f\circ \phi^{-1})'(\phi(p))$$ denotes the Jacobian of $$f\circ \phi^{-1}$$ at $$\phi(p)$$ which is the $$1 \times n$$-matrix $$\left(\dfrac{\partial(f\circ \phi^{-1})}{\partial x_i}(\phi(p))\right)$$ and $$(\phi \circ \gamma)'(0)$$ denotes the column vector $$\left(x_i'(0)\right)$$. Matrix multiplication gives the desired result.