Computing the gradient of the function $\psi(u) = f[ \phi(u)]$ 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.
 A: This is not an answer, but doesn't fit easily into comments. I find your notation a bit confusing, for example, I would write $t_x = \frac{\partial \phi_x(u)}{\partial u}$ instead of what you have above.
The componentwise expression of the chain rule in the comments above is as follows (note that $\psi = f \circ \phi$):
Let $f(x,y,z) = \begin{bmatrix} f_x(x,y,z) \\ f_y(x,y,z) \\ f_z(x,y,z) \end{bmatrix}$, $\phi(u) = \begin{bmatrix} \phi_x(u) \\ \phi_y(u) \\ \phi_z(u) \end{bmatrix}$, $\psi(u) = \begin{bmatrix} f_x(\phi_x(u), \phi_y(u), \phi_z(u)) \\ f_y(\phi_x(u), \phi_y(u), \phi_z(u)) \\ f_z(\phi_x(u), \phi_y(u), \phi_z(u)) \end{bmatrix}$.
Then using the chain rule componentwise you have (I have suppressed the arguments to simplify):
$$
\begin{bmatrix}
\frac{\partial \psi_x}{\partial u} \\
\frac{\partial \psi_y}{\partial u} \\
\frac{\partial \psi_z}{\partial u}
\end{bmatrix}
=
\begin{bmatrix} \frac{\partial f_x}{\partial x}\frac{\partial \phi_x}{\partial u} + \frac{\partial f_x}{\partial y}\frac{\partial \phi_y}{\partial u} + \frac{\partial f_x}{\partial z}\frac{\partial \phi_z}{\partial u} \\
\frac{\partial f_y}{\partial x}\frac{\partial \phi_x}{\partial u} + \frac{\partial f_y}{\partial y}\frac{\partial \phi_y}{\partial u} + \frac{\partial f_y}{\partial z}\frac{\partial \phi_z}{\partial u}  \\ \frac{\partial f_z}{\partial x}\frac{\partial \phi_x}{\partial u} + \frac{\partial f_z}{\partial y}\frac{\partial \phi_y}{\partial u} + \frac{\partial f_z}{\partial z}\frac{\partial \phi_z}{\partial u}  \end{bmatrix} =
\begin{bmatrix} \frac{\partial f_x}{\partial x} & \frac{\partial f_x}{\partial y} & \frac{\partial f_x}{\partial z} \\
\frac{\partial f_y}{\partial x} &
\frac{\partial f_y}{\partial y} &
\frac{\partial f_y}{\partial z} &  \\
\frac{\partial f_z}{\partial x} &
\frac{\partial f_z}{\partial y} &
\frac{\partial f_z}{\partial z} &
\end{bmatrix}
\begin{bmatrix}
\frac{\partial \phi_x}{\partial u} \\
\frac{\partial \phi_y}{\partial u} \\
\frac{\partial \phi_z}{\partial u}
\end{bmatrix}.
$$
The above can be written more succinctly as $D \psi(u) = Df(\phi(u)) \, D\phi(u)$.
The vector $t$ is given by $t = \begin{bmatrix}
\frac{\partial \phi_x}{\partial u} \\
\frac{\partial \phi_y}{\partial u} \\
\frac{\partial \phi_z}{\partial u}
\end{bmatrix}$, and similarly for $t'$ ($\phi$ replaced by $\psi$).
