1. **Problem statement:** We need to find explicit expressions for the partial derivatives $\frac{\partial g_1}{\partial x}$ and $\frac{\partial g_2}{\partial y}$ where $g(x,y) = (g_1(x,y), g_2(x,y))$ satisfies the system $$F(x,y,u,v) = (z,w) = (0,0)$$ with $$z = x^2 - y^2 + u^2 - v^3 + 4$$ $$w = 2xy + y^2 + 3u^4 - 2v^2 + 8$$ and $g(x,y) = (u,v)$ implicitly defined by $F(x,y,g(x,y)) = (0,0)$. 2. **Formula and method:** By the Implicit Function Theorem, the derivatives of $g$ satisfy $$D_{(u,v)}F(x,y,u,v) \cdot Dg(x,y) = -D_{(x,y)}F(x,y,u,v)$$ where $$D_{(u,v)}F = \begin{bmatrix} \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \\ \frac{\partial w}{\partial u} & \frac{\partial w}{\partial v} \end{bmatrix}, \quad D_{(x,y)}F = \begin{bmatrix} \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} \end{bmatrix}$$ and $$Dg = \begin{bmatrix} \frac{\partial g_1}{\partial x} & \frac{\partial g_1}{\partial y} \\ \frac{\partial g_2}{\partial x} & \frac{\partial g_2}{\partial y} \end{bmatrix}$$ We want $\frac{\partial g_1}{\partial x}$ and $\frac{\partial g_2}{\partial y}$, i.e., the $(1,1)$ and $(2,2)$ entries of $Dg$. 3. **Calculate partial derivatives of $F$:** - For $z$: $$\frac{\partial z}{\partial u} = 2u$$ $$\frac{\partial z}{\partial v} = -3v^2$$ $$\frac{\partial z}{\partial x} = 2x$$ $$\frac{\partial z}{\partial y} = -2y$$ - For $w$: $$\frac{\partial w}{\partial u} = 12u^3$$ $$\frac{\partial w}{\partial v} = -4v$$ $$\frac{\partial w}{\partial x} = 2y$$ $$\frac{\partial w}{\partial y} = 2x + 2y$$ 4. **Write matrices:** $$D_{(u,v)}F = \begin{bmatrix} 2u & -3v^2 \\ 12u^3 & -4v \end{bmatrix}$$ $$D_{(x,y)}F = \begin{bmatrix} 2x & -2y \\ 2y & 2x + 2y \end{bmatrix}$$ 5. **Solve for $Dg$:** $$Dg = - (D_{(u,v)}F)^{-1} D_{(x,y)}F$$ 6. **Inverse of $D_{(u,v)}F$:** Calculate determinant: $$\det = (2u)(-4v) - (-3v^2)(12u^3) = -8uv + 36 u^3 v^2 = 4uv(-2 + 9 u^2 v)$$ Inverse: $$ (D_{(u,v)}F)^{-1} = \frac{1}{\det} \begin{bmatrix} -4v & 3v^2 \\ -12 u^3 & 2u \end{bmatrix} $$ 7. **Compute $Dg$ entries:** $$Dg = - \frac{1}{\det} \begin{bmatrix} -4v & 3v^2 \\ -12 u^3 & 2u \end{bmatrix} \begin{bmatrix} 2x & -2y \\ 2y & 2x + 2y \end{bmatrix}$$ Calculate each entry: - First row, first column (for $\frac{\partial g_1}{\partial x}$): $$(-4v)(2x) + (3v^2)(2y) = -8 v x + 6 v^2 y$$ - Second row, second column (for $\frac{\partial g_2}{\partial y}$): $$(-12 u^3)(-2y) + (2u)(2x + 2y) = 24 u^3 y + 4 u x + 4 u y$$ 8. **Final expressions:** $$\frac{\partial g_1}{\partial x} = - \frac{-8 v x + 6 v^2 y}{\det} = \frac{8 v x - 6 v^2 y}{\det}$$ $$\frac{\partial g_2}{\partial y} = - \frac{24 u^3 y + 4 u x + 4 u y}{\det} = - \frac{24 u^3 y + 4 u x + 4 u y}{\det}$$ where $$\det = 4 u v (-2 + 9 u^2 v)$$ **Summary:** $$\boxed{\begin{cases} \frac{\partial g_1}{\partial x} = \frac{8 v x - 6 v^2 y}{4 u v (-2 + 9 u^2 v)} \\ \frac{\partial g_2}{\partial y} = - \frac{24 u^3 y + 4 u x + 4 u y}{4 u v (-2 + 9 u^2 v)} \end{cases}}$$ These formulas give the partial derivatives of $g_1$ and $g_2$ in terms of $x,y,u,v$ as requested.