Given a function $u(\textbf{t}) = \begin{bmatrix}c&a \cdot tx+b \cdot ty+d\end{bmatrix}$ and $\textbf{t}=\begin{bmatrix}tx\\ty\end{bmatrix}$, I need to find: $d u(\textbf{t})/d\textbf{t}$.
My understanding from chain rule is that: $$\frac{d u(\textbf{t})}{d\textbf{t}} = \frac{\partial u(\textbf{t})}{\partial t_x} \cdot \frac{dt_x}{d\textbf{t}} + \frac{\partial u(\textbf{t})}{\partial t_y} \cdot \frac{dt_y}{d\textbf{t}}$$
The issue I have is that I'm not sure how to find $\frac{dt_x}{d\textbf{t}}$ and $\frac{dt_y}{d\textbf{t}}$. I can find $ \frac{d\textbf{t}}{dt_x}$ and $\frac{d\textbf{t}}{dt_y}$, but I'm not sure how to find the inverse.
After doing some research I think I've solved my issue. Since $u(\textbf{t}): \mathbb{R}^2 \rightarrow \mathbb{R}^2$, then $\frac{du(\textbf{t})}{d\textbf{t}}$ will be a jacobian. So the correct form is: $$\frac{du(\textbf{t})}{d\textbf{t}}=\begin{bmatrix}\frac{\partial u(\textbf{t})_1}{\partial(t_x)} & \frac{\partial u(\textbf{t})_1}{\partial(t_y)} \\ \frac{\partial u(\textbf{t})_2}{\partial(t_x)} & \frac{\partial u(\textbf{t})_2}{\partial(t_y)}\end{bmatrix}$$ So I believe that the final result will be: $$\frac{du(\textbf{t})}{d\textbf{t}}=\begin{bmatrix}0 & 0 \\ a & b\end{bmatrix}$$
Source: http://www.cs.huji.ac.il/~csip/tirgul3_derivatives.pdf