Pind the following derivatives. 7 of 8 \( z_{s} \) and \( z_{t} \), where \( z=7 x y-2 x^{2} y, x=2 s+5 t \), and \( y=2 s-5 t \) \( \frac{\partial s}{\partial z}=< \) (Type an expression using s and \( t \) as the variables.) 0.63 of 1 \( \frac{\partial z}{\partial y}=7 x-2 x^{2} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{\partial y}{\partial s}=2 \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial x}{\partial t}=5 \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial y}{\partial t}=-5 \) (Type an expression using \( s \) and \( t \) as the variables.) \( z_{s}=\square \) (Type an expression using \( s \) and \( t \) as the variahloc)

To find the partial derivatives \( z_{s} \) and \( z_{t} \), we start by using the chain rule. 1. The expression for \( z \) is \( z = 7xy - 2x^2y \). 2. We will first calculate \( z_{s} \): \[ z_{s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \] Calculating \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \): \[ \frac{\partial z}{\partial x} = 7y - 4xy \] \[ \frac{\partial z}{\partial y} = 7x - 2x^2 \] Using \( \frac{\partial y}{\partial s} = 2 \) and \( \frac{\partial x}{\partial s} = 2 \): \[ z_{s} = (7y - 4xy)(2) + (7x - 2x^2)(2) \] Thus, consolidating gives us: \[ z_s = 2(7y - 4xy) + 2(7x - 2x^2) = 14y - 8xy + 14x - 4x^2 \] 2. Next, for \( z_{t} \): \[ z_{t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \] Using \( \frac{\partial y}{\partial t} = -5 \) and \( \frac{\partial x}{\partial t} = 5 \): \[ z_{t} = (7y - 4xy)(5) + (7x - 2x^2)(-5) \] Thus: \[ z_t = 5(7y - 4xy) - 5(7x - 2x^2) = 35y - 20xy - 35x + 10x^2 \] In summary: \[ z_{s} = 14y - 8xy + 14x - 4x^2 \] \[ z_{t} = 35y - 20xy - 35x + 10x^2 \]