Part 10 of 11 Find the following derivatives. Express your answer in terms of the independent variables. \( w_{s} \) and \( w_{t} \), where \( w=\frac{x-z}{3 y+z}, x=3 s+4 t, y=s t \), and \( z=3 s-4 t \) os 0 of 1 (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial z}{\partial s}=3 \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial x}{\partial t}=4 \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial y}{\partial t}=s \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial z}{\partial t}=-4 \) (Type an expression using \( s \) and \( t \) as the variables.) \( w_{s}=\square \) (Type an expression using \( s \) and \( t \) as the variables.)

We start with \[ w=\frac{x-z}{3y+z}, \] where \[ x=3s+4t,\quad y=st,\quad z=3s-4t. \] Our goal is to compute the partial derivative of \(w\) with respect to \(s\), denoted \(w_s\). We use the quotient rule which gives \[ w_s = \frac{\frac{\partial}{\partial s}(x-z)(3y+z)- (x-z)\frac{\partial}{\partial s}(3y+z)}{(3y+z)^2}. \] 1. Compute \(\frac{\partial}{\partial s}(x-z)\). Since \[ x=3s+4t\quad \text{and}\quad z=3s-4t, \] we have \[ \frac{\partial x}{\partial s} = 3 \quad \text{and} \quad \frac{\partial z}{\partial s} = 3. \] Thus, \[ \frac{\partial}{\partial s}(x-z)= 3-3 = 0. \] 2. Compute \(\frac{\partial}{\partial s}(3y+z)\). Note that \[ y=st \quad \text{implies} \quad \frac{\partial y}{\partial s}=t, \] and as before, \(\frac{\partial z}{\partial s}=3;\) therefore, \[ \frac{\partial}{\partial s}(3y+z)= 3\frac{\partial y}{\partial s}+ \frac{\partial z}{\partial s} = 3t+3 = 3(t+1). \] 3. Substitute into the quotient rule formula: \[ w_s = \frac{0\cdot(3y+z) - (x-z)\cdot 3(t+1)}{(3y+z)^2} = -\frac{3(t+1)(x-z)}{(3y+z)^2}. \] 4. Express \(x-z\) and \(3y+z\) in terms of \(s\) and \(t\). We have \[ x-z = (3s+4t) - (3s-4t)=8t. \] Also, \[ 3y+z = 3(st) + (3s-4t)= 3st+3s-4t. \] 5. Plug these back to obtain \[ w_s = -\frac{3(t+1)(8t)}{(3st+3s-4t)^2} = -\frac{24t(t+1)}{(3st+3s-4t)^2}. \] Thus, the derivative is \[ w_{s} = -\frac{24t(t+1)}{(3st+3s-4t)^2}. \]