We are given that $z = e^x + 2y$, where $x = \frac{s}{t}$ and $y = \frac{t}{s}$. We are tasked with finding $\frac{\partial z}{\partial s}$ using the chain rule.

Step 1: Apply the chain rule

To compute $\frac{\partial z}{\partial s}$, we use the chain rule to account for the dependencies of $z$ on $x$ and $y$, and their dependencies on $s$ and $t$. The chain rule states:

$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}$

Step 2: Compute the partial derivatives

• First, let's compute $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$:

$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(e^x + 2y) = e^x$ $\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(e^x + 2y) = 2$

• Next, compute $\frac{\partial x}{\partial s}$ and $\frac{\partial y}{\partial s}$:

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

Step 3: Substitute these into the chain rule

Now we substitute the partial derivatives into the chain rule formula:

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

Step 4: Final Expression

Substitute $x = \frac{s}{t}$ back into the equation:

$\frac{\partial z}{\partial s} = \frac{e^{s/t}}{t} - \frac{2t}{s^2}$

This is the final expression for $\frac{\partial z}{\partial s}$.

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Would you like more details on any specific step?

Here are some related questions to explore:

1. How do you compute partial derivatives of exponential functions?
2. What is the general form of the chain rule in multivariable calculus?
3. How would you apply the same method to find $\frac{\partial z}{\partial t}$?
4. How do changes in $s$ affect $z$ when $t$ is held constant?
5. How do we interpret the partial derivatives in this context?

Tip: Always keep track of dependencies when applying the chain rule in multivariable calculus!