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The chain rule is a fundamental concept in calculus used to differentiate composite functions. When dealing with a function like \( f(x, y) = e^{-x^2 - y^2} \), it's clear that there are layers of functions involved. There is the exponential function on the outside, and the polynomial \( -x^2 - y^2 \) on the inside.

When applying the chain rule, we differentiate the outer function and multiply it by the derivative of the inner function. For partial derivatives, this means we treat one variable as a constant and differentiate with respect to the other. For example, to find \( \frac{\partial f}{\partial x} \), we first focus on the exponential function's derivative, which remains the exponential itself, and then multiply it by the derivative of \( -x^2 \) with respect to \( x \), which is \( -2x \).

This process results in:
\[ \frac{\partial f}{\partial x} = -2xe^{-x^2 - y^2} \] Similarly, to find \( \frac{\partial f}{\partial y} \), the polynomial \( y^2 \) is differentiated to give \( -2y \). The chain rule is used frequently in multivariable calculus due to the intricacy of functions involving several variables.

The product rule is employed when differentiating functions that are multiplied together. Its importance becomes especially clear during the calculation of second-order derivatives. Consider the function \( f(x, y) = -2xe^{-x^2 - y^2} \). Here, you have a product of two functions: \(-2x\) and \(e^{-x^2 - y^2} \).

According to the product rule, when differentiating a product, you must differentiate each component separately and sum the results. Hence, for \( \frac{\partial^2 f}{\partial x^2} \):

• First, differentiate \(-2x\) as if \(e^{-x^2 - y^2} \) were a constant.
• Then, differentiate \(e^{-x^2 - y^2} \) with respect to \(x\), using the chain rule as above.

Adding the results gives us:\[ \frac{\partial^2 f}{\partial x^2} = -2e^{-x^2 - y^2} + 4x^2e^{-x^2 - y^2} = (4x^2 - 2)e^{-x^2 - y^2} \]
The same logic applies to the \(y\) derivative, where the product of \(-2y\) and \(e^{-x^2 - y^2}\) is differentiated. Mastering the product rule is crucial for handling expressions involving several multiplied components.

Mixed partial derivatives involve taking derivatives of partial derivatives, often with respect to different variables. For instance, \( \frac{\partial^2 f}{\partial x \partial y} \) denotes differentiating first with respect to \( y \) after taking the derivative with respect to \( x \).

This double-differentiation can be somewhat tricky if the functions are complex, but the principle is straightforward: you chain the differentiation process. Using the function \( -2xe^{-x^2 - y^2} \):

• Differentiating \(-2xe^{-x^2 - y^2} \) with respect to \( y \), gives the derivative of the exponential part using the chain rule.
• The result here is: \( 4xye^{-x^2 - y^2} \).

You can also differentiate \(-2ye^{-x^2 - y^2} \) with respect to \( x \) and, interestingly, you'll get the same result. This symmetry in mixed partial derivatives leads us to Clairaut's theorem.

Clairaut's theorem, an important result in calculus, reveals that the order of differentiation for continuous second partial derivatives doesn't matter. In other words, \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \) as long as the function's mixed derivatives are continuous.

This theorem is integral because it simplifies the analysis and calculation of derivatives. For the function \( f(x, y) = e^{-x^2 - y^2} \), both mixed derivatives were calculated:

• \( \frac{\partial^2 f}{\partial x \partial y} = 4xye^{-x^2 - y^2} \)
• \( \frac{\partial^2 f}{\partial y \partial x} = 4xye^{-x^2 - y^2} \)

They are equivalent, showcasing Clairaut's theorem in action. This symmetry is not just a coincidence but a profound result of the function's continuity. Understanding this theorem can save computation time and aids in problem-solving by offering a predictable outcome when computing mixed derivatives.