We are asked to find the partial derivatives $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$, given:

$z = e^{(x + 5y)}, \quad x = \frac{s}{t}, \quad y = \frac{t}{s}$

Step 1: Differentiate $z$ with respect to $s$

To find $\frac{\partial z}{\partial s}$, we apply the chain rule. First, express $z$ as:

$z = e^{(x + 5y)}$

Now, we need to find:

$\frac{\partial z}{\partial s} = \frac{d}{ds} \left( e^{x + 5y} \right)$

Using the chain rule, we have:

$\frac{\partial z}{\partial s} = e^{(x + 5y)} \left( \frac{\partial}{\partial s} (x + 5y) \right)$

Now, find $\frac{\partial}{\partial s} (x + 5y)$:

$\frac{\partial}{\partial s} (x + 5y) = \frac{\partial x}{\partial s} + 5 \frac{\partial y}{\partial s}$

Since $x = \frac{s}{t}$ and $y = \frac{t}{s}$, we differentiate:

$\frac{\partial x}{\partial s} = \frac{1}{t}, \quad \frac{\partial y}{\partial s} = -\frac{t}{s^2}$

Thus, we have:

$\frac{\partial}{\partial s} (x + 5y) = \frac{1}{t} + 5 \left( -\frac{t}{s^2} \right)$

$\frac{\partial}{\partial s} (x + 5y) = \frac{1}{t} - \frac{5t}{s^2}$

Therefore, the derivative of $z$ with respect to $s$ is:

$\frac{\partial z}{\partial s} = e^{(x + 5y)} \left( \frac{1}{t} - \frac{5t}{s^2} \right)$

Step 2: Differentiate $z$ with respect to $t$

Next, we apply the chain rule to find $\frac{\partial z}{\partial t}$:

$\frac{\partial z}{\partial t} = e^{(x + 5y)} \frac{\partial}{\partial t} (x + 5y)$

Now, differentiate $x + 5y$ with respect to $t$:

$\frac{\partial}{\partial t} (x + 5y) = \frac{\partial x}{\partial t} + 5 \frac{\partial y}{\partial t}$

We differentiate $x = \frac{s}{t}$ and $y = \frac{t}{s}$ with respect to $t$:

$\frac{\partial x}{\partial t} = -\frac{s}{t^2}, \quad \frac{\partial y}{\partial t} = \frac{1}{s}$

Thus:

$\frac{\partial}{\partial t} (x + 5y) = -\frac{s}{t^2} + 5 \left( \frac{1}{s} \right)$

$\frac{\partial}{\partial t} (x + 5y) = -\frac{s}{t^2} + \frac{5}{s}$

Therefore, the derivative of $z$ with respect to $t$ is:

$\frac{\partial z}{\partial t} = e^{(x + 5y)} \left( -\frac{s}{t^2} + \frac{5}{s} \right)$

Final Results:

1. $\frac{\partial z}{\partial s} = e^{(x + 5y)} \left( \frac{1}{t} - \frac{5t}{s^2} \right)$
2. $\frac{\partial z}{\partial t} = e^{(x + 5y)} \left( -\frac{s}{t^2} + \frac{5}{s} \right)$

Would you like further clarification or details? Here are some related questions for additional practice:

1. How do you apply the product rule alongside the chain rule?
2. Can you compute the second partial derivatives $\frac{\partial^2 z}{\partial s^2}$ and $\frac{\partial^2 z}{\partial t^2}$?
3. What happens if you substitute specific values for $s$ and $t$?
4. How does the chain rule extend to functions with more variables?
5. How does the chain rule change when working with implicit functions?

Tip: Always keep track of the inner functions when applying the chain rule, especially when the variables are nested.