Find the following derivatives. Express your answer in terms of the independent variables. \( w_{s} \) and \( w_{t} \), where \( w=\frac{4 x-4 z}{3 y+4 z}, x=3 s+t, y=3 s t \), and \( z=3 s-t \) \( w_{s}=-\frac{8 t(9 t+12)}{(9 s t+12 s-4 t)^{2}} \) (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 the derivative \( w_t \), we will apply the quotient rule to the function \( w = \frac{4x - 4z}{3y + 4z} \). We need to compute both the numerator and denominator derivatives with respect to \( t \). First, let’s calculate the partial derivatives: - For the numerator \( 4x - 4z \): - Differentiating \( 4x \) with respect to \( t \) gives: \[ 4 \frac{dx}{dt} = 4 \cdot 1 = 4 \] - Differentiating \( -4z \) with respect to \( t \): \[ -4 \frac{dz}{dt} = -4 \cdot (-1) = 4 \] So, the total derivative of the numerator is \( 4 + 4 = 8 \). - For the denominator \( 3y + 4z \): - Differentiating \( 3y \) with respect to \( t \) gives: \[ 3 \frac{dy}{dt} = 3 \cdot 3s = 9s \] - Differentiating \( 4z \) with respect to \( t \) gives: \[ 4 \frac{dz}{dt} = 4 \cdot (-1) = -4 \] So, the total derivative of the denominator is \( 9s - 4 \). Now we can apply the quotient rule: \[ w_t = \frac{(3y + 4z)(8) - (4x - 4z)(9s - 4)}{(3y + 4z)^2}. \] Now substitute \( x = 3s + t \), \( y = 3st \), and \( z = 3s - t \): \[ \begin{align*} 9s - 4 & = 3(3st) + 4(3s-t) \\ & = 9s + 12s - 4t \\ & = 21s - 4t. \\ \end{align*} \] Combining these: \[ w_t = \frac{(21s - 4t)(8) - (12s)(9s - 4)}{(21s - 4)^2}. \] Thus, the expression for \( w_t \) is: \[ w_t = \frac{(21s - 4 t)8 - (12s)(9s - 4)}{(21s - 4)^2}. \] Hence: \[ w_t = \frac{(168s - 32t - 108s + 48)}{(21s - 4)^2}. \] Thus the final form is: \[ w_t = \frac{(60s - 32t + 48}{(21s - 4)^2}. \]