Find the derivative of G(x,y) using Jacobian matrix

Question

$$G(x,y)=(g_1(x,y),g_2(x,y))$$ $$g_2(x,y)=g_1(x^3-xy^2,x^y-y^3)$$ $$g_1(1,0)=1,\quad D_1g_1(1,0)=2, D_2g_1(1,0)=1$$

Find the Jacobian matrix G'(1,0)

Answer 1

One way* we can define the Jacobian matrix for $G$ is

$$J = \begin{bmatrix} \frac{\partial g_1}{\partial x} & \frac{\partial g_1}{\partial y} \\ \frac{\partial g_2}{\partial x} & \frac{\partial g_2}{\partial y} \end{bmatrix} \; \; .$$

Since we have that $$\;\;g_2(x,y)=g_1(x^3-xy^2,x^y-y^3)\;\;$$ we can view this with a coordinate transformation $\;\;g_2(x,y)=g_1(u,v)\;\;$ where $u = x^3-xy^2$ and $v=x^y-y^3.$

By the chain rule, we then have that

$$\frac{\partial g_2}{\partial x} = \frac{\partial }{\partial x}g_1(u,v) =\frac{\partial g_1}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial g_1}{\partial v}\frac{\partial v}{\partial x}$$

$$\text{and}$$

$$\frac{\partial g_2}{\partial y} = \frac{\partial }{\partial y}g_1(u,v) =\frac{\partial g_1}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial g_1}{\partial v}\frac{\partial v}{\partial y}$$

$$\text{thus}$$

$$J = \begin{bmatrix} \frac{\partial g_1}{\partial x} & \frac{\partial g_1}{\partial y} \\ \frac{\partial g_1}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial g_1}{\partial v}\frac{\partial v}{\partial x} & \frac{\partial g_1}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial g_1}{\partial v}\frac{\partial v}{\partial y} \end{bmatrix} \; \; .$$

I'll leave you to work out what the partials of $u$ and $v$ are. All that is left is for you to plug in the values for $(x,y)$ and the derivatives of $g_1$ and $g_2$. Note that (as a hint) you have that

$$D_1g_1(1,0)=2$$

thus

$$\frac{\partial g_1}{\partial u}(u=1,v=0) = 2$$

$$\text{and}$$

$$\frac{\partial g_1}{\partial x}(x=1,y=0) = 2 \; \; .$$

*One could also define this as $J^T$

$\textbf{EDIT: } $ to address @Physor's comment.

Something that needs to be made more clear is that you will have to evaluate $u(x=1,y=0)$ and $v(x=1,y=0)$. These values are what you should plug in for the partials of $g_1$ and $g_2$ with respect to $u$ and $v$. Using an explicit example, evaluating $\frac{\partial g_1}{\partial u}$ at $(x,y) = (1,0)$ means you are computing

$$D_1 g_1\big(u(1,0),v(1,0)\big) \; .$$

It should also be noted that since $v(1,0)\neq 0$ under the OP's current definition of the problem, the problem cannot be currently solved with the information provided. If the OP needs to correct $v(x,y)$ to something where $v(1,0)= 0$, we know that $$D_1 g_1\big(u(1,0),v(1,0)\big)=2 \; .$$

Answer 2

$$ DG(x,y) = \begin{pmatrix} \partial_1 g_1 & \partial_2 g_1 \\ \partial_1 g_2 & \partial_2 g_2 \\ \end{pmatrix} \\ $$ we have $g_2(x,y)=g_1(x^3-xy^2,x^y-y^3)$, then \begin{align} Dg_2(x,y) &= \begin{pmatrix} \partial_1 g_2 & \partial_2 g_2 \\ \end{pmatrix} \\&= \begin{bmatrix} (3x^2-y^2)\partial_1 g_1 +(yx^{y-1})\partial_2 g_1 & (-2xy)\partial_1 g_1 +(\ln(x)x^y-3y^2)\partial_2 g_1 \end{bmatrix} \end{align} Substitute in $DG(x,y)$ $$ DG(x,y) = \begin{pmatrix} \partial_1 g_1 & \partial_2 g_1 \\ (3x^2-y^2)\partial_1 g_1 +(yx^{y-1})\partial_2 g_1 & (-2xy)\partial_1 g_1 +(\ln(x)x^y-3y^2)\partial_2 g_1 \end{pmatrix} \\ $$

There might be a mistake since I didn't use $g_1(1,0)=1$ anywhere, see also my comment under the question