# Why does this smooth function's derivative vanish at the point $c$?

I'm reading my notes and I can't figure out this statement. Let $(U,\varphi$) be a coordinate chart for an $n$ dimensional manifold $M$ around $x$ with coordinates $(x_1,\dots, x_n)$. Let $f\in C^\infty(M)$ and let $\phi=f\varphi^{-1}$. Consider $$F=f-\sum \frac{\partial\phi}{\partial x_i}(\varphi(c))x_i$$ Then why does the derivative of $F$ vanish at $c$?

I think I am confused about how to differentiate the sum.

## 2 Answers

It suffices to write it in coordinates. Let's use the chart $(U,\varphi)$ that we already have. We have \begin{align*} {\rm d}F_c &= {\rm d}\left(f-\sum_i \frac{\partial \phi}{\partial x^i}(\varphi(c))x^i \right)_c \\ &= {\rm d}f_c - \sum_i \frac{\partial \phi}{\partial x^i}(\varphi(c)){\rm d}(x^i)_c \\ &= \sum_i \frac{\partial \phi}{\partial x^i}(\varphi(c)) {\rm d}x^i- \sum_i \frac{\partial \phi}{\partial x^i}(\varphi(c)) {\rm d}x^i \\ &= 0.\end{align*}Just remember that $\sum_i \frac{\partial \phi}{\partial x^i}(\varphi(c)) {\rm d}x^i$ is precisely the coordinate expression for ${\rm d}f_c$, and, for future reference, that we denote $$\frac{\partial f}{\partial x^i}(c) \stackrel{\cdot}{=} \frac{\partial(f \circ \varphi^{-1})}{\partial x^i}(\varphi(c)) = \frac{\partial \phi}{\partial x^i}(\varphi(c)),$$where the last two derivatives are in $\Bbb R^n$.

Did you try to compute the derivative of $F$ at $c$? The simple way to state it is that you're taking $f$ and removing the first order part of its Taylor polynomial around $c$, and this coincides, up to order $1$ at least, with $f$.

• @jennifer Check the answer by user Ivo Terek for that. – Pedro Tamaroff Feb 19 '17 at 18:52