Derivation in $\mathbb R_{\gt 0} ^2$ with difference quotients My definition of the difference quotient is $f(a+h) = f(a) + Ah +r(h)$ for a linear function A.
Let $f(\begin{pmatrix} x \\y\end{pmatrix}) = \begin{pmatrix} -3x+2+y \\ ln(1+x+2y)\end{pmatrix}$.
$\Longrightarrow f(a+h) - f(a) = \begin{pmatrix} -3x -3h_1 +2+y+h_2 \\ ln(1+x+h_1+2y +2h_2)\end{pmatrix} - \begin{pmatrix} -3x+2+y \\ ln(1+x+2y)\end{pmatrix} =\begin{pmatrix} -3h_1 +h_2 \\ln(1+x+h_1+2y +2h_2)- ln(1+x+2y)\end{pmatrix} = \begin{pmatrix} -3 &1 \\ ? & ? \end{pmatrix} \cdot \begin{pmatrix} h_1 \\ h_2\end{pmatrix} + r(h)$
I know what i have to put for the "?" but how do i get to that?  
 A: To expand on amd's comment, consider using the first order Taylor's formula:
$$\ln(a+t) = \ln(a) + \frac{t}{a} +r(t), \quad \text{where} \quad \lim_{t \rightarrow 0}\frac{r(t)}{t} = 0$$
Let $a = 1+x+2y$ and $t = h_1+2h_2$, then we have
$$\ln(1+x+2y +h_1+2h_2) = \ln(1+x+2y)+\frac{h_1+2h_2}{1+x+2y} + r(h_1+2h_2)$$
But then we have to argue that $\lim_{h \rightarrow \vec0} \frac{r(h_1+h_2)}{\sqrt{h_1^2 +h_2^2}} = 0$. We can do this by using the fact that $|h_1+2h_2| \leq 3\sqrt{h_1^2 +h_2^2}$.
So that the desired linear transformation (for the second component!) is $$T_2(h_1, h_2) = \frac{h_1+2h_2}{1+x+2y}=\begin{bmatrix}
     \frac{1}{1+x+2y}  & \frac{2}{1+x+2y}
\end{bmatrix}\begin{bmatrix}
     h_1 \\
    h_2 \\
\end{bmatrix}
$$

I will then show how to use partial derivative to do that:
Let $f: \Bbb{R^2}_{>0} \rightarrow \Bbb{R}^2$ be a function. We define the directional derivative $f'(a;u)$  as follows:
$$f'(a;u)=\lim_{t \rightarrow 0} \frac{f(a+t u)-f(a)}{t}$$
, where $t \in \Bbb{R}$. We read this as the directional derivative $f$ at $a\ (\in \Bbb{R^2}_{>0})$ at the direction of $u\ (\in \Bbb{R}^2)$. The partial derivative $D_1f(a)= \frac{\partial f}{\partial x}(a): = f'(a;e_1)$.


If $f$ is differentiable at $a$ with total derivative $f'(a)= T$, then for all $u \in \Bbb{R}^2$, $T(u) = f'(a;u)$. 

Proof: If we spell out the definition of total derivative at $a$, we have
$$f(a+tu) = f(a)+tT(u) + r(tu), \quad \text{where} \quad \lim_{|tu| \rightarrow 0}\frac{|r(tu)|}{|tu|} = 0$$
I think you can show that $\lim_{t \rightarrow 0}\frac{|r(tu)|}{t} = 0$ from here, so that we are done.
From what we have defined above, we have $T(e_1) = D_1f(a)$ and $T(e_2)= D_2f(a)$. So that if $u = (u_1, u_2)$, then from linearity of $T$, we have $T(u) = u_1D_1f(a)+ u_2D_2f(a)$.
If $f = (f_1, f_2)$, it follows that $D_1f(a) = (D_1f_1(a), D_1f_2(a))$. Putting all these together,
$$
T(u)=u_1D_1f(a)+ u_2D_2f(a)=u_1
\begin{bmatrix}
    D_1f_1(a)  \\
    D_1f_2(a)
\end{bmatrix}+ u_2\begin{bmatrix}
    D_2f_1(a)  \\
    D_2f_2(a)
\end{bmatrix}=
\begin{bmatrix}
     D_1f_1(a)  & D_2f_1(a) \\
    D_1f_2(a) & D_2f_2(a) \\
\end{bmatrix}
\begin{bmatrix}
     u_1 \\
    u_2 \\
\end{bmatrix}
$$
So the Jacobian $J_f(a)$ is just
\begin{bmatrix}
     D_1f_1(a)  & D_2f_1(a) \\
    D_1f_2(a) & D_2f_2(a) \\
\end{bmatrix}

Let's believe that $f$ is differentiable on $\Bbb{R}^2_{>0}$ for now. Put $f_1(x,y) = -3x+2+y $ and $f_2(x,y) = \ln(1+x+2y)$. From mulitvariable calculus, we have
$$D_1f_1(x,y) = -3, \quad D_2f_1(x,y) = 1$$
$$D_1f_2(x,y) = \frac{1}{1+x+2y}, \quad D_2f_2(x,y) = \frac{2}{1+x+2y}$$
$$J_f(x,y) = \begin{bmatrix}
     -3  & 1 \\
   \frac{1}{1+x+2y} & \frac{2}{1+x+2y}\\
\end{bmatrix}
$$
But actually we have just proved if $f$ is differentiable, then $J_f$ given by that formula. There are functions whose partial derivatives exist, but not differentiable (even discontinuous) at that point. Anyway, we shall not worry too much in this question since the partials are continuous on $\Bbb{R}^2_{>0}$. Continuity of partials is logically equivalent to the function being continuously differentiable.
