Jacobian matrix with two equations Evaluate the Jacobian for:
$$f(x,y)=(x^2+x+y, yx+x^2)$$
at the point $(1,2)$.
 A: The Jacobian matrix, see the simple example, is given by:
$$J(x, y) = \left(
\begin{array}{cc}
 2 x+1 & 1 \\
 2 x+y & x \\
\end{array}
\right)$$
So, we have:
$$J(1,2) = \left(
\begin{array}{cc}
 3 & 1 \\
 4 & 1 \\
\end{array}
\right)$$
A: The Jacobian of $f=(f_1,f_2):\mathbb{R}^2\to \mathbb{R}^2$ at $(x,y)$ (where $f_1,f_2:\mathbb{R}\to \mathbb{R}$ are the (first and second, respectively) coordinate functions of $f$) is defined by the rule:
$J_{(x,y)}(f)=\begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{pmatrix}$. 
In your question, $f_1(x,y)=x^2+x+y$ and $f_2(x,y)=yx+x^2$. Also, as a sample computation check that $\frac{\partial f_1}{\partial x}=2x+1$. Now can you calculate the matrix above? Once you've done that, then you can substitute your point $x=1$ and $y=2$ to obtain:
$J_{(1,2)}(f)$ - a $2\times 2$ matrix of real numbers.
According to the computation $\frac{\partial f_1}{\partial x}=2x+1$, the $(1,1)$-entry of this matrix will be 3.
Hope this helps! Please let me know if you're still stuck on something and I'm very happy to help.
