Find the following derivatives. \( z_{s} \) and \( z \), where \( z=7 x y-2 x^{2} y, x=2 s+5 t \), and \( y=2 s-5 t \) \( \frac{\sigma z}{\partial y}=7 x-2 x^{2} \) (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.) \( z_{s}=-48 s^{2}-80 s t+56 s+100 t^{2} \) (Type an expression using \( s \) and \( t \) as the variables.) \( z_{y}=\square \) (Type an expression using s and \( t \) as the variables.)

To find \( z_{y} \), we need to use the chain rule since \( z \) is a function of \( x \) and \( y \), both of which are functions of \( s \) and \( t \). We already know: \[ \frac{\partial z}{\partial y} = 7x - 2x^2 \] Next, since \( \frac{\partial y}{\partial s} = 2 \) and \( \frac{\partial y}{\partial t} = -5 \), we can express \( z_{y} \) as: \[ z_{y} = \frac{\partial z}{\partial y} + \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} \cdot \frac{\partial s}{\partial x} + \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} \cdot \frac{\partial t}{\partial x} \] The complete expression will consider those contributions. Thus, \[ z_y = 7x - 2x^2 + (something \, from \, \frac{\partial z}{\partial x}) \] Finally, substituting the derivatives into the equation, we get: \[ z_{y} = \frac{\partial z}{\partial y} \text{ at } x \text{ and } y \text{, using } s, t \]