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

To find the derivative \( w_s \), we can use the chain rule. First, we need to differentiate \( w \) with respect to \( s \): 1. **Express \( w \)** in terms of \( s \) and \( t \): \[ w = \frac{4(3s+t) - 4(3s-t)}{3(3s t) + 4(3s-t)} = \frac{12s + 4t - 12s + 4t}{9st + 12s - 4t} = \frac{8t}{9st + 12s - 4t} \] 2. **Applying the Quotient Rule**: Let \( u = 8t \) and \( v = 9st + 12s - 4t \). Then, \[ w = \frac{u}{v} \] The derivative \( w_s \) using the quotient rule is given by: \[ w_s = \frac{u'v - uv'}{v^2} \] 3. **Calculating \( u' \)**: Since \( u = 8t \), we have: \[ u' = 0 \quad (\text{since } t \text{ is treated as a constant with respect to } s) \] 4. **Calculating \( v' \)**: For \( v = 9st + 12s - 4t \): \[ v' = 9t + 12 \] 5. **Plugging into the derivative formula**: \[ w_s = \frac{0 \cdot v - 8t(9t + 12)}{v^2} = -\frac{8t(9t + 12)}{(9st + 12s - 4t)^2} \] Now we have: \[ w_s = -\frac{8t(9t + 12)}{(9st + 12s - 4t)^2} \] Thus, the expression for \( w_s \) is: \[ w_{s} = -\frac{8t(9t + 12)}{(9st + 12s - 4t)^2} \]