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+## Calculating Derivatives $dx(u)$ and $dy(u)$ Over a Triangular Finite Element
+
+
+<img width="200" align="left" src="https://github.com/user-attachments/assets/124852e5-3169-4436-8fa9-b90bddfc9972"/>
+
+To calculate the derivatives of a function $u$ with respect to $x$ and $y$ ( denoted as $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, or more simply $dx(u)$, $dy(u)$ ) over a triangular finite element, we follow these steps:
+
+### 1. Define the Triangle and the P1 Basis Functions
+
+In a P1 (linear) finite element, the function $u$ is assumed to be linear over each triangle. 
+Let the vertices of the triangle be $\mathbf{v}_1 = (x_1, y_1)$, $\mathbf{v}_2 = (x_2, y_2)$, and $\mathbf{v}_3 = (x_3, y_3)$,
+with corresponding function values $u_1 = u(\mathbf{v}_1)$, $u_2 = u(\mathbf{v}_2)$, and $u_3 = u(\mathbf{v}_3)$.
+
+The function $u(x, y)$ can be expressed as:
+
+$$u(x, y) = a_0 + a_1 x + a_2 y$$
+
+where $a_0$, $a_1$, and $a_2$ are coefficients to be determined.
+
+### 2. Set Up the System of Equations
+
+We solve for the coefficients $a_0$, $a_1$, and $a_2$ using the known values of $u$ at the triangle's vertices:
+
+$$
+\begin{aligned}
+u_1 &= a_0 + a_1 x_1 + a_2 y_1  \\
+u_2 &= a_0 + a_1 x_2 + a_2 y_2  \\
+u_3 &= a_0 + a_1 x_3 + a_2 y_3 
+\end{aligned}
+$$
+
+This system can be written in matrix form:
+
+$$
+\begin{pmatrix}
+1 & x_1 & y_1  \\
+1 & x_2 & y_2  \\
+1 & x_3 & y_3 
+\end{pmatrix}
+\begin{pmatrix}
+a_0 \\
+a_1 \\
+a_2 
+\end{pmatrix}=
+\begin{pmatrix}
+u_1 \\
+u_2 \\
+u_3
+\end{pmatrix}
+$$
+
+
+Solve this system to get $a_0$, $a_1$, and $a_2$.
+
+### 3. Calculate the Derivatives $dx(u)$ and $dy(u)$
+
+- The derivative $\frac{\partial u}{\partial x}$ is simply $a_1$. This is because the partial derivative of $u(x, y)$ with respect to $x$ is:
+
+$$
+\frac{\partial u}{\partial x} = a_1
+$$
+
+Therefore, $dx(u) = a_1$.
+
+- The derivative $\frac{\partial u}{\partial y}$ is simply $a_2$. This is because the partial derivative of $u(x, y)$ with respect to $y$ is:
+
+$$
+\frac{\partial u}{\partial y} = a_2
+$$
+
+Therefore, $dy(u) = a_2$.
+
+### 4. Shortcut Using the Area Formula (2D case)
+
+For a 2D triangular element (i.e., $z = 0$), the coefficients $a_1$ and $a_2$ (which give the derivatives with respect to $x$ and $y$, respectively) can be directly computed using the area of the triangle and the coordinates of its vertices:
+
+$$
+a_1 = \frac{1}{2A} \left( u_1(y_2 - y_3) + u_2(y_3 - y_1) + u_3(y_1 - y_2) \right)
+$$
+
+$$
+a_2 = \frac{1}{2A} \left( u_1(x_3 - x_2) + u_2(x_1 - x_3) + u_3(x_2 - x_1) \right)
+$$
+
+where $A$ is the area of the triangle, given by:
+
+$$
+A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
+$$
+
+This method directly gives the derivatives $dx(u)$ and $dy(u)$ without solving a system of equations.
+
+## TO DO ##
+ADD CODE EXAMPLE