Find the following derivatives. \( z_{s} \) and \( z_{\mathrm{t}} \), where \( z=9 x y-3 x^{2} y, x=3 s+5 t \), and \( y=3 s-5 t \) \( z_{s}=-243 s^{2}+162 s-270 s t+225 t^{2} \) (Type an expression using \( s \) and \( t \) as the variables.) \( z_{t}=\square \) (Type an expression using \( s \) and \( t \) as the variables.)

To find the derivatives \( z_s \) and \( z_t \), we first need to use the chain rule. Starting with the given function \( z = 9xy - 3x^2y \), we need to compute \( z_s \) and \( z_t \) after substituting \( x = 3s + 5t \) and \( y = 3s - 5t \). 1. **Finding \( z_t \):** We know from your question that: \[ z_s = -243s^2 + 162s - 270st + 225t^2 \] Now, let's find \( z_t \). Using the chain rule: \[ z_t = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \] First, compute \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \): \[ \frac{\partial z}{\partial x} = 9y - 6xy \] \[ \frac{\partial z}{\partial y} = 9x - 3x^2 \] Now calculate \( \partial x/\partial t \) and \( \partial y/\partial t \): \[ \frac{\partial x}{\partial t} = 5 \] \[ \frac{\partial y}{\partial t} = -5 \] Now substitute the \( x \) and \( y \) back into the derivatives: So, \[ z_t = (9(3s - 5t) - 6(3s + 5t)(3s - 5t)) \cdot 5 + (9(3s + 5t) - 3(3s + 5t)^2)(-5) \] Calculating everything gives: \[ z_t = [-135s + 180t - 30(9s^2 - 15st - 15st + 25t^2)] \cdot 5 + [27s + 45t - 3(9s^2 + 30st + 25t^2)] \cdot (-5) \] Putting it all together, we simplify and find: \[ z_t = -243s^2 + 162s + 270st - 225t^2 \] So, the final solution for \( z_t \) is: \( z_t = -243 s^2 + 162 s + 270 s t - 225 t^2 \)