ind the following derivatives. \( z_{s} \) and \( z_{\text {}} \), 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.) \( \frac{\partial z}{\partial y}=5 e^{3 x+5 y} \) (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 the derivatives \( \frac{\partial z}{\partial x} \), \( \frac{\partial x}{\partial s} \), \( \frac{\partial z}{\partial y} \), and \( \frac{\partial y}{\partial s} \), we will use the chain rule and the given expressions for \( z \), \( x \), and \( y \). Given: \( z = e^{3x+5y} \) \( x = 8st \) \( y = 6s + 5t \) 1. Find \( \frac{\partial z}{\partial x} \): Using the chain rule, we have: \( \frac{\partial z}{\partial x} = 3e^{3x+5y} \) 2. Find \( \frac{\partial x}{\partial s} \): Differentiating \( x = 8st \) with respect to \( s \), we get: \( \frac{\partial x}{\partial s} = 8t \) 3. Find \( \frac{\partial z}{\partial y} \): Using the chain rule, we have: \( \frac{\partial z}{\partial y} = 5e^{3x+5y} \) 4. Find \( \frac{\partial y}{\partial s} \): Differentiating \( y = 6s + 5t \) with respect to \( s \), we get: \( \frac{\partial y}{\partial s} = 6 \) Therefore, the derivatives are: \( \frac{\partial z}{\partial x} = 3e^{3x+5y} \) \( \frac{\partial x}{\partial s} = 8t \) \( \frac{\partial z}{\partial y} = 5e^{3x+5y} \) \( \frac{\partial y}{\partial s} = 6 \)