Find the total differential. \[ w=\frac{10 x+y}{4 z-8 y} \]

To find the total differential of the function \( w = \frac{10x + y}{4z - 8y} \), we can use the quotient rule. The total differential \( dw \) is given by: \[ dw = \frac{\partial w}{\partial x} dx + \frac{\partial w}{\partial y} dy + \frac{\partial w}{\partial z} dz \] First, we compute the partial derivatives. Let \( u = 10x + y \) and \( v = 4z - 8y \). Using the quotient rule: 1. **Partial derivative with respect to \( x \)**: \[ \frac{\partial w}{\partial x} = \frac{v \cdot \frac{\partial u}{\partial x} - u \cdot \frac{\partial v}{\partial x}}{v^2} \] Here, \( \frac{\partial u}{\partial x} = 10 \) and \( \frac{\partial v}{\partial x} = 0 \), thus: \[ \frac{\partial w}{\partial x} = \frac{(4z - 8y)(10) - (10x + y)(0)}{(4z - 8y)^2} = \frac{40z - 80y}{(4z - 8y)^2} \] 2. **Partial derivative with respect to \( y \)**: \[ \frac{\partial w}{\partial y} = \frac{v \cdot \frac{\partial u}{\partial y} - u \cdot \frac{\partial v}{\partial y}}{v^2} \] Here, \( \frac{\partial u}{\partial y} = 1 \) and \( \frac{\partial v}{\partial y} = -8 \): \[ \frac{\partial w}{\partial y} = \frac{(4z - 8y)(1) - (10x + y)(-8)}{(4z - 8y)^2} = \frac{4z - 8y + 80x + 8y}{(4z - 8y)^2} = \frac{4z + 80x}{(4z - 8y)^2} \] 3. **Partial derivative with respect to \( z \)**: \[ \frac{\partial w}{\partial z} = \frac{v \cdot \frac{\partial u}{\partial z} - u \cdot \frac{\partial v}{\partial z}}{v^2} \] Here, \( \frac{\partial u}{\partial z} = 0 \) and \( \frac{\partial v}{\partial z} = 4 \): \[ \frac{\partial w}{\partial z} = \frac{(4z - 8y)(0) - (10x + y)(4)}{(4z - 8y)^2} = \frac{-4(10x + y)}{(4z - 8y)^2} \] Now, putting it all together, the total differential \( dw \) is: \[ dw = \left(\frac{40z - 80y}{(4z - 8y)^2} \right) dx + \left(\frac{4z + 80x}{(4z - 8y)^2} \right) dy + \left(\frac{-4(10x + y)}{(4z - 8y)^2}\right) dz \] In simplified form, we get: \[ dw = \frac{1}{(4z - 8y)^2} \left( (40z - 80y) dx + (4z + 80x) dy - 4(10x + y) dz \right) \]