# Find a jacobian matrix of a function

### Here's the problem:

Let $$F:R^4 \to R^2, g:R^2\to R^2, h:R\to R$$ all differentiable, such that:

$$g(x_0,y_0) = (2,3), h(x_0) = 1, h(y_0) = 4$$.

Suppose that $$F(h(x),g(x,y),h(y)) = (0,0), \forall(x,y) \in R^2$$.

The following Jacobian Matrices are given:

$$JF(1,2,3,4) = \bigg[\matrix{1\\-2}\matrix{7\\0}\matrix{0\\-5}\matrix{2\\5}\bigg], Jh(x_0) = \big[\matrix{14}\big], Jh(y_0) = \big[\matrix{10}\big]$$.

Calculate the jacobian matrix of $$g$$ in the point $$(x_0,y_0)$$.

### What I Know

I'm really lost here... I know how to calculate a Jacobian Matrix, given a function, but I don't know how to start here or what I can use to solve it. I just need a tip to start.

• Number 1. Find the function Jan 7 at 23:28
• @1613585 can you help me with that? Initially I thought about the Jacobian matrix of composition of functions but I didn't find anything that could help me. Jan 8 at 0:35

Rewriting some equations of your problem for convenience

$$h(x) = h_1(x), \, h(y) = h_2(y)$$

$$g(x,y) = \left[\begin{matrix}g_1(x,y)\\g_2(x,y)\end{matrix}\right]$$

$$f(h(x),g(x,y),h(y)) = f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\begin{matrix}f_1(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\\f_2(h_1(x),g_1(x,y),g_2(x,y),h_2(y))\end{matrix}\right] = \left[\begin{matrix}0\\0\end{matrix}\right]$$

$$J_f(h_1(x),g_1(x,y),g_2(x,y),h_2(y)) = \left[\matrix{\frac{\partial f_1}{\partial h_1}\\\frac{\partial f_2}{\partial h_1}}\matrix{\frac{\partial f_1}{\partial g_1}\\\frac{\partial f_2}{\partial g_1}}\matrix{\frac{\partial f_1}{\partial g_2}\\\frac{\partial f_2}{\partial g_2}}\matrix{\frac{\partial f_1}{\partial h_2}\\\frac{\partial f_2}{\partial h_2}}\right]$$

$$J_g(x,y) = \left[\matrix{\frac{\partial g_1}{\partial x}\\\frac{\partial g_2}{\partial x}}\matrix{\frac{\partial g_1}{\partial y}\\\frac{\partial g_2}{\partial y}}\right]$$

$$J_{h_1}(x) = \frac{\partial h_1}{\partial x}$$

$$J_{h_2}(y) = \frac{\partial h_2}{\partial y}$$

From the chain rule, we know that for $$i={1,2}$$

$$\frac{\partial f_i}{\partial x} = 0 = \frac{\partial f_i}{\partial h_1} \frac{\partial h_1}{\partial x} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial x} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial x}$$

$$\frac{\partial f_i}{\partial y} = 0 = \frac{\partial f_i}{\partial h_2} \frac{\partial h_2}{\partial y} + \frac{\partial f_i}{\partial g_1} \frac{\partial g_1}{\partial y} + \frac{\partial f_i}{\partial g_2} \frac{\partial g_2}{\partial y}$$

It is given that

$$J_f(h(x_0),g_1(x_0,y_0),g_2(x_0,y_0),h(y_0)) = \bigg[\matrix{1\\-2}\matrix{7\\0}\matrix{0\\-5}\matrix{2\\5}\bigg], J_{h_1}(x_0) = \big[\matrix{14}\big], J_{h_2}(y_0) = \big[\matrix{10}\big]$$

So it is possible to write the system

$$14 + 7 \frac{\partial g_1}{\partial x} (x_0,y_0) = 0$$

$$20 + 7 \frac{\partial g_1}{\partial y} (x_0,y_0) = 0$$

$$-28 -5 \frac{\partial g_2}{\partial x} (x_0,y_0) = 0$$

$$50 - 5 \frac{\partial g_2}{\partial y} (x_0,y_0) = 0$$