Find the following derivatives. Express your answer in terms of the independent variables. \[ w_{s} \text { and } w_{t} \text {, where } w=\frac{x-z}{3 y+z}, x=3 s+4 t, y=s t \text {, and } z=3 s-4 t \] \( \frac{\partial w}{\partial x}=\square \) (Type an expression usirty \( x, y \), and \( z \) as the variables.)

To find the derivative \( \frac{\partial w}{\partial x} \), we first note that \( w = \frac{x - z}{3y + z} \). The next step is to compute \( \frac{\partial w}{\partial x} \). Using the quotient rule for derivatives, where \( u = x - z \) and \( v = 3y + z \): \[ \frac{\partial w}{\partial x} = \frac{v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}}{v^2} \] Calculating \( \frac{\partial u}{\partial x} \): \[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x - z) = 1 \] Next, we calculate \( \frac{\partial v}{\partial x} \): Since \( y = st \), we have: \[ \frac{\partial y}{\partial x} = 0 \] And \( z = 3s - 4t \) gives: \[ \frac{\partial z}{\partial x} = 0 \] Thus, \[ \frac{\partial v}{\partial x} = 3 \cdot 0 + 0 = 0 \] Substituting these into the quotient rule expression for the derivative: \[ \frac{\partial w}{\partial x} = \frac{(3y + z)(1) - (x - z)(0)}{(3y + z)^2} = \frac{3y + z}{(3y + z)^2} \] This simplifies to: \[ \frac{\partial w}{\partial x} = \frac{1}{3y + z} \] Thus, the answer is: \[ \frac{\partial w}{\partial x} = \frac{1}{3y + z} \]