Given are a parametric surface $S(u,v) = \big( x(u,v), y(u,v), z(u,v) \big)$ and a scalar function $f(u,v)$, both defined on the same domain $\Omega \subset \mathbb{R}^2$. As such, we can associate every point on the surface $S(u,v)$ with a scalar value $f(u,v)$. Throughout, it is assumed that the surface is regular and the function $f$ differentiable.
Out of curiosity, I'm interested in deriving an expression for the tangential directional derivative of the function $f$ as seen from the surface $S(u,v)$, which I'll denote $\tilde{f}$. That is, considering a point $p = S(u,v)$ and a nearby point $q = S(u+ha, v+hb)$ with $h$ variable and for some (domain) direction $w = \begin{pmatrix}a\\b\end{pmatrix}$, what is the result of
$$\lim_{h \to 0} \frac{\tilde{f}(q) - \tilde{f}(p)}{\|q-p\|}.$$
Below I've written down my approach — the question is whether the result is correct (I do have some doubts about it, as explained below).
As the numerator is the difference of two function values, it is equivalent to $f(u+ha, v+hb) - f(u,v)$. Substituting the expressions for $p$ and $q$, the denominator is $\| S(u+ha, v+hb) - S(u,v) \|$. From the Taylor expansion, we know that $S(u+ha, v+hb) \approx S(u,v) + ha \frac{\partial S}{\partial u}(u,v) + hb \frac{\partial S}{\partial v}(u,v)$ for small $h$, which in the limit becomes an equivalence. As such, in the limit the denominator represents the length of a tangent vector I'll denote $h \tilde{w}$. Introducing the $3 \times 2$ Jacobian matrix $J = \big(\frac{\partial S}{\partial u}, \frac{\partial S}{\partial v}\big)$, we can express this tangent vector as $h \tilde{w} = h J w$.
We can now re-write the above as
$$\lim_{h \to 0} \frac{\tilde{f}(q) - \tilde{f}(p)}{\| q - p \|} = \lim_{h \to 0} \frac{f(u+ha, v+hb) - f(u,v)}{h} \frac{h}{\| q - p \|} = D_w f(u,v) \frac{1}{\| \tilde{w} \|} = \frac{\nabla f \cdot w}{\| \tilde{w} \|},$$
where $D_w f(u,v)$ is the directional derivative of $f(u,v)$ in the direction of $w$. Next, with $\tilde {w} = J w$, we have $w = J^+ \tilde{w}$, with $J^+$ the (left) Moore-Penrose inverse. That is, $J^+ = (J^TJ)^{-1} J^T$. Introducing the first fundamental form (i.e. the metric tensor) $g = J^TJ$, we can write $J^+ = g^{-1} J^T$. Therefore, we have
$$\frac{\nabla f \cdot w}{\| \tilde{w} \|} = \frac{ \left(\nabla f\right)^T g^{-1} J^T \tilde{w}}{\| \tilde{w} \|} = \left( J g^{-1} \nabla f \right) \cdot \frac{\tilde{w}}{\| \tilde{w} \|},$$
which follows from transposing (recall that $g$ is symmetric and therefore $g^{-1}$ is as well).
On the one hand, this looks promising, as $J g^{-1} \nabla f$ matches what I've seen described as the tangential gradient (also referred to as the surface gradient) in the literature. Therefore, like the ordinary directional derivative, the tangential directional derivative appears to be the dot product of a gradient and a direction vector. On the other hand, I'm a bit sceptical about this vector appearing normalised. Is this a result of using this specific approach, or did I make a mistake/wrong assumption somewhere along the way?
