Find the following derivatives. \( z_{s} \) and \( z_{t} \), where \( z=7 x y-2 x^{2} y, x=2 s+5 t \), and \( y=2 s-5 t \) \( \frac{\partial z}{\partial x}=7 y-4 x y \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{\partial x}{\partial s}=\square \) (Type an expression using s and \( t \) as the variables.)

To find the derivatives \( z_s \) and \( z_t \) from the given function \( z = 7xy - 2x^2y \), we first need to compute \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) using the chain rule. First, we can express the partial derivatives in terms of \(s\) and \(t\): 1. Use the given derivative \( \frac{\partial z}{\partial x} = 7y - 4xy \). 2. Calculate \( \frac{\partial z}{\partial y} \): - \( \frac{\partial z}{\partial y} = 7x - 2x^2 \). 3. Find \( \frac{\partial x}{\partial s} = 2 \) and \( \frac{\partial x}{\partial t} = 5 \). 4. Find \( \frac{\partial y}{\partial s} = 2 \) and \( \frac{\partial y}{\partial t} = -5 \). Now apply the chain rule to find \( z_s \) and \( z_t \): 1. \( z_s = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \) - Thus, \( z_s = (7y - 4xy)(2) + (7x - 2x^2)(2) \). 2. \( z_t = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \) - Therefore, \( z_t = (7y - 4xy)(5) + (7x - 2x^2)(-5) \). Finally, we can express the results: - \( z_s = 2(7y - 4xy) + 2(7x - 2x^2) \) - \( z_t = 5(7y - 4xy) - 5(7x - 2x^2) \) Now, substituting for \( \frac{\partial x}{\partial s} \): \( \frac{\partial x}{\partial s} = 2 \) So, to summarize: 1. \( z_s = 2(7y - 4xy) + 2(7x - 2x^2) \) 2. \( z_t = 5(7y - 4xy) - 5(7x - 2x^2) \) 3. \( \frac{\partial x}{\partial s} = 2 \)