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

We start with \[ z = e^{3x+5y}, \] with \[ x = 8st \quad \text{and} \quad y = 6s + 5t. \] **Step 1. Compute the derivatives of \( z \) with respect to \( x \) and \( y \).** Since \[ z = e^{3x+5y}, \] by the chain rule we have \[ \frac{\partial z}{\partial x} = 3e^{3x+5y} \quad \text{and} \quad \frac{\partial z}{\partial y} = 5e^{3x+5y}. \] **Step 2. Compute \( z_s \) using the chain rule.** The chain rule gives: \[ z_s = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}. \] We now calculate the necessary derivatives: - From \( x = 8st \): \[ \frac{\partial x}{\partial s} = 8t. \] - From \( y = 6s + 5t \): \[ \frac{\partial y}{\partial s} = 6. \] Thus, \[ z_s = \left(3e^{3x+5y}\right)(8t) + \left(5e^{3x+5y}\right)(6). \] Simplify: \[ z_s = 24t\, e^{3x+5y} + 30e^{3x+5y}. \] **Step 3. Compute \( z_t \) using the chain rule.** Similarly, \[ z_t = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}. \] Now compute the derivatives with respect to \( t \): - From \( x = 8st \): \[ \frac{\partial x}{\partial t} = 8s. \] - From \( y = 6s + 5t \): \[ \frac{\partial y}{\partial t} = 5. \] Thus, \[ z_t = \left(3e^{3x+5y}\right)(8s) + \left(5e^{3x+5y}\right)(5). \] Simplify: \[ z_t = 24s\, e^{3x+5y} + 25e^{3x+5y}. \] **Step 4. Write the derivative \(\frac{\partial z}{\partial x}\) in terms of \(x\) and \(y\).** We already found that \[ \frac{\partial z}{\partial x} = 3e^{3x+5y}, \] which is the expression in terms of \(x\) and \(y\). **Final Answers:** \[ z_s = 24t\, e^{3x+5y} + 30e^{3x+5y}, \] \[ z_t = 24s\, e^{3x+5y} + 25e^{3x+5y}, \] \[ \frac{\partial z}{\partial x} = 3e^{3x+5y}. \]