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On page 250 of Modern Differential Geometry of Curves and Surfaces with Mathematica, Second Edition, by Alfred Gray https://books.google.com/books?id=-LRumtTimYgC&lpg=PP1&pg=PA250#v=onepage&q&f=false

"Lemma 11.3. (The Chain Rule) Let $g_{1},\dots,g_{k}:\mathbb{R}^{n}\rightarrow\mathbb{R}$ and $F:\mathbb{R}^{k}\rightarrow\mathbb{R}$ be differentiable functions and let $\boldsymbol{v}_{\boldsymbol{p}}\in\mathbb{R}_{\boldsymbol{p}}^{n}$ where $\boldsymbol{p}\in\mathbb{R}^{n}$. Write $f=F\left(g_{1},\dots,g_{k}\right)$. Then

$\boldsymbol{v}_{\boldsymbol{p}}\left[f\right]=\underset{j=1}{\overset{k}{\sum}}\frac{\partial F}{\partial u_{j}}\left(g_{1}\left(\boldsymbol{p}\right),\dots,g_{k}\left(\boldsymbol{p}\right)\right)\boldsymbol{v}_{\boldsymbol{p}}\left[g_{j}\right]$."

I contend that the partial differentiation should be with respect to $g_{j}$ instead of $u_{j}$. That is

$\boldsymbol{v}_{\boldsymbol{p}}\left[f\right]=\underset{j=1}{\overset{k}{\sum}}\frac{\partial F}{\partial g_{j}}\left(g_{1}\left(\boldsymbol{p}\right),\dots,g_{k}\left(\boldsymbol{p}\right)\right)\boldsymbol{v}_{\boldsymbol{p}}\left[g_{j}\right]$.

Am I correct?

The lemma is stated a bit differently in the third edition, but the expression in question is of the same form. See page 267, Lemma 9.6.

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    Your expression is not an equality or something, is there a typo? – 2017-02-02
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    @JohnMa Indeed. I have corrected the expressions. Thank you for taking the time to look at this. – 2017-02-02

1 Answers 1

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Gray is correct. But my alternative formulation is also correct.

$\frac{\partial F}{\partial u_{j}}$ is the rate of change of F with respect to the coordinate value $u_{j}$.

$\frac{\partial F}{\partial g_{j}}$ is the rate of change of F with respect to the coordinate value which is given as a function $g_{j}$.

The two denote the same value with different connotations.