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Partial derivatives help us understand how a function changes when we alter just one variable while keeping others constant. For first-order partial derivatives, we focus on the rates of change in two directions for the given function.
In our example function, \( z = x e^{x/y^2} \), we first need to find these derivatives with respect to the variables \( x \) and \( y \).

The first step is to differentiate the function treating other variables as constants.

• The derivative of \( z \) concerning \( x \), denoted as \( z_x \), involves simply applying basic differentiation rules since \( y \) is treated as a constant.
  
• On the other hand, finding \( z_y \) means using the chain rule, as changing \( y \) impacts the exponential term as well as \( y^2 \) in the denominator.

These first-order partial derivatives are crucial for understanding the initial behavior of the function in response to small changes in \( x \) or \( y \).

The chain rule is an essential tool in calculating derivatives when dealing with composite functions, where one function is nested inside another. In the context of partial derivatives, it helps us differentiate expressions that involve functions of more complex variables.

For instance, in the function \( z = x e^{x/y^2} \), applying the chain rule is necessary while computing its derivative with respect to \( y \). The expression \( e^{x/y^2} \) requires us to view it as an exponential function with an internal function \( x/y^2 \).

Using the chain rule:

• The derivative of \( e^{x/y^2} \) with respect to \( y \) involves taking the derivative of the entire exponential function and then multiplying it by the derivative of its inside component, \( x/y^2 \), with respect to \( y \).
• This yields terms comprising both exponential and rational components, showcasing the chain rule's ability to handle complications from nested expressions.

Mastering the chain rule allows us to better tackle multivariable calculus problems like those involving partial derivatives.

When we calculate partial derivatives of different variables in sequence, we get mixed partial derivatives. For a function \( z(x, y) \), these are the derivatives of the form \( \frac{\partial^2 z}{\partial x \partial y} \) and \( \frac{\partial^2 z}{\partial y \partial x} \).

In the context of our example, it means differentiating first with respect to one variable, say \( x \), and then differentiating the result with respect to another, \( y \), or vice versa.

Clairaut's theorem on the equality of mixed partials states that for functions meeting certain conditions, the mixed derivatives \( z_{xy} \) and \( z_{yx} \) should be equal.

In our problem:

• After calculating \( z_x \) as an intermediate step, differentiating again with respect to \( y \) gives us \( z_{xy} \).
  
• Similarly, calculating \( z_y \) first and then differentiating with respect to \( x \) provides \( z_{yx} \).
  
• Both \( z_{xy} \) and \( z_{yx} \) end up being the same, which reaffirms Clairaut's theorem, making this a helpful checkpoint to confirm your calculations.

Understanding mixed partial derivatives deepens our insight into how multivariable functions behave.

The product rule is used when differentiating an expression that represents a product of two functions. When dealing with partial derivatives, it’s crucial when functions of multiple variables are multiplied together.

Consider the initial function \( z = x e^{x/y^2} \), where the product of \( x \) and \( e^{x/y^2} \) requires the product rule application.

Using the product rule entails:

• First, take the derivative of the first function \( x \), which is simply 1, while leaving \( e^{x/y^2} \) constant.
  
• Next, take the derivative of the second function \( e^{x/y^2} \) while keeping \( x \) constant. This step generally requires the chain rule due to the nature of the exponential function.
  
• Combine these results: the derivative of the first function times the second plus the first function times the derivative of the second.

Using the product rule in partial derivatives allows handling complex expressions in a structured way, simplifying the calculation process.