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 \( z=3 s-4 t \) (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.) \( w_{s}=-\frac{24 t(t+1)}{(3 s t+3 s-4 t)^{2}} \) (Type an expression using \( s \) and \( t \) as the variables.) \( w_{t}=\square \) (Type an expression using \( s \) and \( t \) as the variables.)

To find \( w_t \), we can use the chain rule to differentiate \( w \) with respect to \( t \): \[ w_t = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial t} \] Now we need the partial derivatives \( \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \) and \( \frac{\partial w}{\partial z} \). Using the quotient rule, we can compute them: \[ \frac{\partial w}{\partial x} = \frac{(3y + z)}{(3y + z)^2} \] \[ \frac{\partial w}{\partial y} = \frac{-(x - z)}{(3y + z)^2} \cdot 3 = -\frac{3(x - z)}{(3y + z)^2} \] \[ \frac{\partial w}{\partial z} = \frac{(x - z)}{(3y + z)^2} \cdot 1 = \frac{x - z}{(3y + z)^2} \] Now substituting the given partial derivatives and the expressions for derivatives of \( x, y, \) and \( z \) with respect to \( t \): \[ w_t = \left(\frac{3y + z}{(3y + z)^2}\right)(4) + \left(-\frac{3(x - z)}{(3y + z)^2}\right)(s) + \left(\frac{x - z}{(3y + z)^2}\right)(-4) \] Combining these terms, we have: \[ w_t = \frac{4(3y + z) - 3s(x - z) - 4(x - z)}{(3y + z)^2} \] To sum it all up, now we compute the values: 1. \( 3y + z = 3(s t) + (3s - 4t) = 3st + 3s - 4t \) 2. \( x - z = (3s + 4t) - (3s - 4t) = 8t \) Finally, plugging these back into the expression for \( w_t \): \[ w_t = \frac{4(3st + 3s - 4t) - 3s(8t) - 4(8t)}{(3st + 3s - 4t)^2} \] \[ = \frac{12st + 12s - 16t - 24st - 32t}{(3st + 3s - 4t)^2} \] \[ = \frac{-12st + 12s - 48t}{(3st + 3s - 4t)^2} \] Thus, the answer for \( w_t \) expressed in terms of \( s \) and \( t \) is: \[ w_t = \frac{12s - 12st - 48t}{(3st + 3s - 4t)^2} \]