Assume that \( F(x, y, z(x, y))=0 \) implicitly defines \( z \) as a differentiable function of \( x \) and \( y \). The partial derivatives of \( z \) are \( \frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}} \) and \( \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}} \). Evaluate \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) for \( x y z+2 x+3 y-5 z=0 \). \( \frac{\partial z}{\partial x}= \)

To evaluate \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) for the equation \( F(x, y, z) = xy z + 2x + 3y - 5z = 0 \), we first need to compute the partial derivatives \( F_x \), \( F_y \), and \( F_z \). 1. Compute the partial derivatives: - \( F_x = y z + 2 \) - \( F_y = x z + 3 \) - \( F_z = xy - 5 \) 2. Using the formula: - \( \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \) and - \( \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} \) Now substitute the values: - For \( \frac{\partial z}{\partial x} \): \[ \frac{\partial z}{\partial x} = -\frac{y z + 2}{xy - 5} \] Thus, the final answer for \( \frac{\partial z}{\partial x} \) is: \[ \frac{\partial z}{\partial x} = -\frac{y z + 2}{xy - 5} \]