This question is based on section 6 of the paper Kriging and splines with derivative information.
A parametric curve $\phi(u)$ in three dimensions is deformed by the function $f$ to a new curve $\psi(u) = f[ \phi(u)]$.
As a result, for any parameter $u$, a point $\mathbf{s} = \phi(u)$ is mapped to a new point $\mathbf{s}^\prime = \psi(u)$ and the gradient $\mathbf{t} = \dot{\phi}(u)$ is mapped to a new gradient $\mathbf{t}^\prime = \dot{\psi}(u)$.
I am trying to express the gradient $\mathbf{t}^\prime$ in terms of $\mathbf{t}$. Here is my attempt.
Starting from the expression $\psi(u) = f[ \phi(u)]$, by the chain rule, for the $x$ component of $\psi(u)$
$\frac{\partial \psi}{\partial x} = \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x} \mbox{ } (\ast)$
where $\psi$ and $\phi$ are taken to mean $\psi(u)$ and $\phi(u)$ respectively. Because
$\frac{\partial \phi}{\partial x} = \mathbf{t}_x \mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ } \frac{\partial \psi}{\partial x} = \mathbf{t}_x^\prime$
Equation $(\ast)$ becomes
$\mathbf{t}_x^\prime = \frac{\partial f}{\partial \phi} \mathbf{t}_x$
By a similar argument,
$\mathbf{t}_y^\prime = \frac{\partial f}{\partial \phi} \mathbf{t}_y \mbox{ }\mbox{ }\mbox{ }\mbox{ and }\mbox{ }\mbox{ }\mbox{ } \mathbf{t}_z^\prime = \frac{\partial f}{\partial \phi} \mathbf{t}_z$
The problem is that I don't know what to make of $\frac{\partial f}{\partial \phi}$. From what I understand from the paper this term is supposed to be a component of the gradient $\nabla f$. Unfortunately, I can't explain why this is so.