How do I find/approximate the Jacobian of this ODE system? $$x_1'=(x_2-x_4)x_5-x_1+8$$
$$x_2'=(x_3-x_5)x_1-x_2+8$$
$$x_3'=(x_4-x_1)x_2-x_3+8$$
$$x_4'=(x_5-x_2)x_3-x_4+8$$
$$x_5'=(x_1-x_3)x_4-x_5+8$$
I am asked for either an exact or approximate solution. I understand that $F$ is a transformation from $\mathbb{R}^5$ to $\mathbb{R}^5$. Hence I have to find the 5 by 5 matrix $\left(\frac{dF_1}{dx}\frac{dF_2}{dx}\frac{dF_3}{dx}\frac{dF_4}{dx}\frac{dF_5}{dx}\right)$. But I have no way of getting the exact form of $F$, so I am not sure what to do. 
 A: I am assuming that this Jacobian "$F$" is $\frac{df}{dx}$ where $x'=f(x)$.
I will compute a few elements of this matrix to give you an idea.
$$ x' = \begin{bmatrix}x_1' \\ x_2' \\ x_3' \\ x_4' \\ x_5' \end{bmatrix} = f(x) = \begin{bmatrix}f_1(x) \\ f_2(x) \\ f_3(x) \\ f_4(x) \\ f_5(x) \end{bmatrix} = \begin{bmatrix}(x_2-x_4)x_5-x_1+8 \\ (x_3-x_5)x_1-x_2+8 \\ (x_4-x_1)x_2-x_3+8 \\ (x_5-x_2)x_3-x_4+8 \\ (x_1-x_3)x_4-x_5+8 \end{bmatrix}$$
$$F = \frac{df}{dx}
= \begin{bmatrix} \frac{df_1}{dx_1} & \frac{\partial f_1}{\partial x_2} & \frac{\partial f_1}{\partial x_3} & \frac{\partial f_1}{\partial x_4} & \frac{\partial f_1}{\partial x_5} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \frac{\partial f_2}{\partial x_3} & \frac{\partial f_2}{\partial x_4} & \frac{\partial f_2}{\partial x_5} \\ \frac{\partial f_3}{\partial x_1} & \frac{\partial f_3}{\partial x_2} & \frac{\partial f_3}{\partial x_3} & \frac{\partial f_3}{\partial x_4} & \frac{\partial f_3}{\partial x_5} \\ \frac{\partial f_4}{\partial x_1} & \frac{\partial f_4}{\partial x_2} & \frac{\partial f_4}{\partial x_3} & \frac{\partial f_4}{\partial x_4} & \frac{\partial f_4}{\partial x_5} \\ \frac{\partial f_5}{\partial x_1} & \frac{\partial f_5}{\partial x_2} & \frac{\partial f_5}{\partial x_3} & \frac{\partial f_5}{\partial x_4} & \frac{\partial f_5}{\partial x_5} \end{bmatrix}
= \begin{bmatrix} -1 & x_5 & 0 & -x_5 & (x_2-x_4) \\ (x_3-x_5) & \cdot & \cdot & \cdot & \cdot \\ -x_2 & \cdot & \cdot & \cdot & \cdot \\ 0 & \cdot & \cdot & \cdot & \cdot \\ x_4 & \cdot & \cdot & \cdot & \cdot \end{bmatrix}$$
