1. **Stating the problem:** We have an implicit function defining pixel intensity $z$ in terms of spatial coordinates $x$ and $y$: $$F(x,y,z) = x^2 z + y \ln(z) - 5 = 0$$ We want to find the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ using the Implicit Function Theorem. 2. **Formula and rules:** By the Implicit Function Theorem, if $F(x,y,z) = 0$ implicitly defines $z$ as a function of $x$ and $y$, then: $$\frac{\partial z}{\partial x} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}, \quad \frac{\partial z}{\partial y} = - \frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$$ 3. **Calculate partial derivatives of $F$:** - $\frac{\partial F}{\partial x} = 2x z$ (since $z$ depends on $x$, but treat $z$ as independent here) - $\frac{\partial F}{\partial y} = \ln(z)$ - $\frac{\partial F}{\partial z} = x^2 + y \frac{1}{z}$ (derivative of $y \ln(z)$ w.r.t. $z$ is $y \frac{1}{z}$) 4. **Write expressions for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$:** $$\frac{\partial z}{\partial x} = - \frac{2 x z}{x^2 + \frac{y}{z}}$$ $$\frac{\partial z}{\partial y} = - \frac{\ln(z)}{x^2 + \frac{y}{z}}$$ 5. **Evaluate at point $(x,y,z) = (2,5,1)$:** - Compute denominator: $$x^2 + \frac{y}{z} = 2^2 + \frac{5}{1} = 4 + 5 = 9$$ - Compute numerator for $\frac{\partial z}{\partial x}$: $$2 x z = 2 \times 2 \times 1 = 4$$ - Compute numerator for $\frac{\partial z}{\partial y}$: $$\ln(1) = 0$$ 6. **Calculate final values:** $$\frac{\partial z}{\partial x} = - \frac{4}{9}$$ $$\frac{\partial z}{\partial y} = - \frac{0}{9} = 0$$ **Final answer:** $$\boxed{\frac{\partial z}{\partial x} = -\frac{4}{9}, \quad \frac{\partial z}{\partial y} = 0}$$