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

To find the derivatives \( z_s \) and \( z_t \), let's start by using the chain rule for partial derivatives given the expressions provided. We know that: \[ z = e^{3x + 5y} \] 1. To find \( z_s \), we apply the chain rule: \[ z_s = \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} + \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} \] From the information given: \[ \frac{\partial z}{\partial x} = 3 e^{3x + 5y} \] \[ \frac{\partial z}{\partial y} = 5 e^{3x + 5y} \] Next, we need to find \( \frac{\partial y}{\partial s} \): Given \( y = 6s + 5t \): \[ \frac{\partial y}{\partial s} = 6 \] Substitute this back into our expression for \( z_s \): \[ z_s = (5 e^{3x + 5y}) \cdot (6) + (3 e^{3x + 5y}) \cdot \left(8t\right) \] Thus: \[ z_s = 30 e^{3x + 5y} + 24t e^{3x + 5y} = (30 + 24t) e^{3x + 5y} \] 2. Now to find \( z_t \): Using the chain rule again: \[ z_t = \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \] From \( y = 6s + 5t \): \[ \frac{\partial y}{\partial t} = 5 \] Then we can find: \[ z_t = (5 e^{3x + 5y}) \cdot 5 = 25 e^{3x + 5y} \] In summary: \[ z_s = (30 + 24t) e^{3x + 5y} \] \[ z_t = 25 e^{3x + 5y} \]