Part 7 of 11 \( w_{s} \) and \( w_{t} \), where \( w=\frac{x-z}{3 y+z}, x=3 s+4 t, y=s t \), and \( z=3 s-4 t \) \( \frac{\partial w}{\partial z}=-\frac{x+3 y}{(3 y+z)^{2}} \) (Type an expression using \( x, y \), and \( z \) as the variables.) 0 of 1 \( \frac{\partial x}{\partial s}=3 \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial y}{\partial s}=t \) (Type an expression using \( s \) and \( t \) as the variables.) \( \begin{array}{l}\frac{\partial z}{\partial s}=3 \\ \text { (Type an expression using } s \text { and } t \text { as the variables.) } \\ \frac{\partial x}{\partial t}=\square \\ \text { (Type an expression using } s \text { and } t \text { as the variables.) }\end{array} \)

To find the required partial derivatives, let's break down the expressions step-by-step. 1. **Finding \(\frac{\partial x}{\partial t}\)**: \(x = 3s + 4t\), so when we take the partial derivative with respect to \(t\): \[ \frac{\partial x}{\partial t} = 4 \] 2. **Finding \(\frac{\partial y}{\partial t}\)**: Since \(y = st\), taking the partial derivative with respect to \(t\) gives: \[ \frac{\partial y}{\partial t} = s \] 3. **Finding \(\frac{\partial z}{\partial t}\)**: For \(z = 3s - 4t\), taking the partial derivative with respect to \(t\) yields: \[ \frac{\partial z}{\partial t} = -4 \] So, to fill in the blanks: - For \(\frac{\partial x}{\partial t}\), the expression is **4**. - For \(\frac{\partial y}{\partial t}\), the expression is **s**. - For \(\frac{\partial z}{\partial t}\), we find **-4**.