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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?

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1 Answer 1

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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.

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