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

To find \( z_s \) and \( z_{t} \), we will use the chain rule and the given relationships: 1. We have \( z = 7xy - 2x^2y \). 2. Given \( x = 2s + 5t \) and \( y = 2s - 5t \). First, let's compute \( z_s \): Using the chain rule: \[ z_s = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \] We already have: \[ \frac{\partial z}{\partial x} = 7y - 4xy \] \[ \frac{\partial z}{\partial y} = 7x - 2x^2 \] \[ \frac{\partial x}{\partial s} = 2 \] We need to find \( \frac{\partial y}{\partial s} \): \[ y = 2s - 5t \implies \frac{\partial y}{\partial s} = 2 \] Now, plug in the values: \[ z_s = (7y - 4xy)(2) + (7x - 2x^2)(2) \] This simplifies to: \[ z_s = 2(7y - 4xy) + 2(7x - 2x^2) \] \[ = 14y - 8xy + 14x - 4x^2 \] Next, we compute \( z_t \): Using the chain rule again: \[ z_t = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \] We need to find \( \frac{\partial x}{\partial t} \) and \( \frac{\partial y}{\partial t} \): \[ \frac{\partial x}{\partial t} = 5 \] \[ \frac{\partial y}{\partial t} = -5 \] Now plug in all the values: \[ z_t = (7y - 4xy)(5) + (7x - 2x^2)(-5) \] \[ = 35y - 20xy - 35x + 10x^2 \] In summary: \[ \frac{\partial y}{\partial s} = 2 \] \[ z_s = 14y - 8xy + 14x - 4x^2 \] \[ z_t = 35y - 20xy - 35x + 10x^2 \]