Use the chain rule to find 
∂z
∂s
 and 
∂z
∂t
.
z = e^(x + 5y),   x = s/t ,   y = t/s

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.

Use the chain rule to find ∂z/∂s and ∂z/∂t. z = e^(x + 5y), x = s/t, y = t/s

Use the chain rule to calculate the partial derivatives ∂z/∂s and ∂z/∂t for z = e^(x + 5y), where x = s/t and y = t/s. This problem involves applying the chain rule to functions of multiple variables and is suitable for undergraduate calculus students.

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