Find the following derivatives. \( \mathrm{z}_{\mathrm{s}} \) and \( \mathrm{z}_{\mathrm{L}} \), where \( \mathrm{z}=e^{3 \mathrm{x}+5 \mathrm{y}}, \mathrm{x}=8 \mathrm{st} \), and \( \mathrm{y}=6 \mathrm{~s}+5 \mathrm{t} \) \[ \overline{\partial \mathrm{s}}=8 \mathrm{t} \] (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.) \[ \frac{\partial y}{\partial s}=\overline{6} \] (Type an expression using \( s \) and \( t \) as the variables.) \[ \frac{\partial x}{\partial t}=8 s \] (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 variables.)

To find \( z_s \), we can use the chain rule. We have: \[ z = e^{3x + 5y} \] Thus, we apply the chain rule: \[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \] Given that \( \frac{\partial z}{\partial x} = 3e^{3x + 5y} \), \( \frac{\partial z}{\partial y} = 5e^{3x + 5y} \), \( \frac{\partial x}{\partial s} = 0 \) (since x does not depend on s), and \( \frac{\partial y}{\partial s} = 6 \), we substitute these into our expression: \[ z_s = 0 + 5 e^{3x + 5y} \cdot 6 = 30 e^{3x + 5y} \] This leads to the expression: \[ z_s = 30 e^{3(8st) + 5(6s + 5t)} \quad (\text{using } x = 8st \text{ and } y = 6s + 5t) \] Next, for \( z_L \), we will seek to derive it using a similar format. The derivative notation \( z_L \) typically denotes \( \frac{\partial z}{\partial l} \) where \( l \) is an independent variable. However, if we take the total derivatives including the relationships we derived: \[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \] Using \( \frac{\partial x}{\partial t} = 8s \) and \( \frac{\partial y}{\partial t} = 5 \): \[ z_L = 3e^{3x + 5y} \cdot 8s + 5e^{3x + 5y} \cdot 5 = (24s + 25) e^{3x + 5y} \] Thus, we provide the expressions as: \[ z_s = 30 e^{24st + 30s + 25t} \] \[ z_L = (24s + 25) e^{24st + 30s + 25t} \]