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 \( x y z+2 x+3 y-5 z=0 \). \( \frac{\partial z}{\partial x}=-\frac{y z+2}{x y-5} \) \( \frac{\partial z}{\partial y}= \)
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To evaluate \( \frac{\partial z}{\partial y} \) for the equation \( F(x, y, z) = xy z + 2x + 3y - 5z = 0 \), we first need to compute the required partial derivatives \( F_x \), \( F_y \), and \( F_z \). Calculating the partial derivatives: 1. \( F_x = y z + 2 \) 2. \( F_y = x z + 3 \) 3. \( F_z = x y - 5 \) Now, using the formula for \( \frac{\partial z}{\partial y} \): \[ \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{x z + 3}{x y - 5} \] So, the final result is: \[ \frac{\partial z}{\partial y} = -\frac{x z + 3}{x y - 5} \]