Second order  directional derivative

You've correctly identified that you need to apply the gradient operator to the expression inside the parentheses, but the calculation of the individual components of this gradient isn't as straightforward as you've written. The confusion seems to be related to the interaction of the partial derivatives and the direction vector.

To correctly calculate the second-order directional derivative, you'll need to apply the directional derivative operator twice. Here's the corrected calculation:

Since $D_uf(x,y)$ is a linear combination of $f_x(x,y)$ and $f_y(x,y)$, you can apply the chain rule to each term separately:

$\frac{\partial}{\partial x} (D_uf) = \frac{\partial}{\partial x} \left( f_x(x,y) \cdot a + f_y(x,y) \cdot b \right)$

$\frac{\partial}{\partial y} (D_uf) = \frac{\partial}{\partial y} \left( f_x(x,y) \cdot a + f_y(x,y) \cdot b \right)$

After these calculations, you'll have the components of the gradient of the first-order directional derivative. Finally, you'll calculate the dot product of this gradient with the direction vector $\vec{u}$.


$\vec{\nabla} \left( D_uf(x,y) \right) \cdot \vec{u}$ =$\begin{bmatrix} \frac{\partial}{\partial x} (D_uf)\\ \frac{\partial}{\partial y} (D_uf)\end{bmatrix} $$\begin{bmatrix} a\\ b\end{bmatrix} $