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

To find \( z_s \), we first need to use the chain rule and substitute \( x \) and \( y \) into the expression for \( z \). Given: \[ z = 9xy - 3x^2y \] Substituting \( x = 3s + 5t \) and \( y = 3s - 5t \): \[ z = 9(3s + 5t)(3s - 5t) - 3(3s + 5t)^2(3s - 5t) \] Now, to derive \( z_s \): 1. Calculate \( \frac{\partial z}{\partial x} = 9y - 6xy \). 2. Calculate \( \frac{\partial z}{\partial y} = 9x - 3x^2 \). 3. Calculate \( \frac{dx}{ds} = 3 \) and \( \frac{dy}{ds} = 3 \). Using the chain rule: \[ z_s = \frac{\partial z}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial z}{\partial y} \cdot \frac{dy}{ds} \] Substituting in our derivatives: \[ z_s = (9y - 6xy)(3) + (9x - 3x^2)(3) \] Now, we can replace \( x \) and \( y \) with their expressions in terms of \( s \) and \( t \) to preserve the variables: \[ z_s = 3(9(3s - 5t) - 6(3s + 5t)(3s - 5t)) + 3(9(3s + 5t) - 3(3s + 5t)^2) \] This leads to: \[ z_s = 27(3s - 5t) + 9(3s + 5t) - 9(3s + 5t)(3s + 5t) \] So, \[ z_s = 27(3s - 5t) + 27(3s + 5t) - 9(3s + 5t)(3s + 5t) \] Final Answer: \[ z_{s} = 27s + 18t - 9(3s + 5t)(3s + 5t) \] In simplified terms, \( z_s \) can be expressed as: \( z_s = 81s + 27t - 9(3s + 5t)(3s + 5t) \) Now, we can also find \( z_t \) in a similar manner. Would you like to proceed with that?