Find the following derivatives. \( z_{s} \) and \( z_{t} \), where \( z=e^{3 x+5 y}, x=8 s t \), and \( y=6 s+5 t \) \( \frac{\partial z}{\partial x}=3 e^{3 x+5 y} \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{\partial x}{\partial s}=8 t \) (Type an expression using \( s \) and \( t \) as the variables.) \( \frac{\partial z}{\partial y}=\square \) (Type an expression using \( x \) and \( y \) as the variables.)

To find the derivatives \( z_{s} \) and \( z_{t} \) where \( z = e^{3x + 5y}, x = 8st, y = 6s + 5t \), we will first apply the chain rule. First, for \( z_{s} \): \[ z_{s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \] We already know: \[ \frac{\partial z}{\partial x} = 3 e^{3x + 5y} \] and \[ \frac{\partial x}{\partial s} = 8t \] Next, we need to determine \( \frac{\partial y}{\partial s} \): \[ \frac{\partial y}{\partial s} = 6 \] Now, we substitute: \[ z_{s} = 3 e^{3x + 5y} \cdot 8t + \frac{\partial z}{\partial y} \cdot 6 \] Next, let’s find \( \frac{\partial z}{\partial y} \): \[ \frac{\partial z}{\partial y} = 5 e^{3x + 5y} \] Now, substitute this expression back into our equation for \( z_{s} \): \[ z_{s} = 3 e^{3x + 5y} \cdot 8t + 5 e^{3x + 5y} \cdot 6 = (24t + 30) e^{3x + 5y} \] Next, we find \( z_{t} \): \[ z_{t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \] First, determine \( \frac{\partial x}{\partial t} \) and \( \frac{\partial y}{\partial t} \): \[ \frac{\partial x}{\partial t} = 8s \] \[ \frac{\partial y}{\partial t} = 5 \] Now we substitute this into the equation for \( z_{t} \): \[ z_{t} = 3 e^{3x + 5y} \cdot 8s + 5 e^{3x + 5y} \cdot 5 = (24s + 25) e^{3x + 5y} \] Final answers are: \[ z_{s} = (24t + 30) e^{3x + 5y} \] \[ z_{t} = (24s + 25) e^{3x + 5y} \] Now, for \( \frac{\partial z}{\partial y} \): \[ \frac{\partial z}{\partial y} = 5 e^{3x + 5y} \]