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Verify that \(z_{x y}=z_{y x}\). $$ z=x^{2} \exp \left(-y^{2}\right) $$

Short Answer

Expert verified
Verified: \(z_{xy} = z_{yx}\).

Step by step solution

01

Understanding the partial derivatives

To verify that \(z_{xy} = z_{yx}\), we need to compute the mixed partial derivatives of \(z\) with respect to \(x\) and \(y\) and see if they are equal.
02

Calculating \(z_x\)

First, calculate the partial derivative of \(z\) with respect to \(x\):\(z = x^2 \, \exp(-y^2)\)Differentiating with respect to \(x\) (treating \(y\) as a constant):\(z_x = 2x \, \exp(-y^2)\)
03

Calculating \(z_{xy}\)

Next, calculate the partial derivative of \(z_x\) with respect to \(y\):\(z_x = 2x \, \exp(-y^2)\)Differentiating with respect to \(y\):\(z_{xy} = 2x \, \frac{d}{dy}[\exp(-y^2)] = 2x \, (-2y) \, \exp(-y^2) = -4xy \, \exp(-y^2)\)
04

Calculating \(z_y\)

Now, calculate the partial derivative of \(z\) with respect to \(y\):\(z = x^2 \, \exp(-y^2)\)Differentiating with respect to \(y\):\(z_y = x^2 \, \frac{d}{dy}[\exp(-y^2)] = x^2 \, (-2y) \, \exp(-y^2) = -2x^2y \, \exp(-y^2)\)
05

Calculating \(z_{yx}\)

Finally, calculate the partial derivative of \(z_y\) with respect to \(x\):\(z_y = -2x^2y \, \exp(-y^2)\)Differentiating with respect to \(x\):\(z_{yx} = -2y \, \frac{d}{dx}[x^2] \, \exp(-y^2) = -4xy \, \exp(-y^2)\)
06

Verification

Both mixed partial derivatives we calculated are equal:\(z_{xy} = -4xy \, \exp(-y^2)\) and \(z_{yx} = -4xy \, \exp(-y^2)\)Therefore, \(z_{xy} = z_{yx}\). This verifies that the mixed partial derivatives are equal as expected.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mixed Partial Derivatives
Mixed partial derivatives refer to the process of differentiating a function multiple times with respect to different variables. In the exercise, we have a function \(z = x^2 \exp(-y^2)\). To find the mixed partial derivatives, we first differentiate with respect to one variable, and then with respect to another. For instance, first find \(z_x\), which is the partial derivative with respect to \(x\), and then differentiate \(z_x\) concerning \(y\) to get \(z_{xy}\). Similarly, find \(z_y\) and then \(z_{yx}\).

These derivatives are crucial in analyzing how a function changes in a multidimensional space. They show how the function responds to small changes in the variables. A common property of mixed partial derivatives (under certain conditions) is that \(z_{xy}\) often equals \(z_{yx}\), which is known as Clairaut's theorem on equality of mixed partial derivatives.
Partial Derivative with Respect to x
Calculating the partial derivative of a function with respect to \(x\) involves differentiating the function while treating the other variables as constants. For our function \(z = x^2 \exp(-y^2)\), the partial derivative with respect to \(x\) is found by focusing on the \(x^2\) term.

We differentiate this term, applying the standard rules of differentiation, resulting in \(z_x = 2x \exp(-y^2)\). This derivative tells us how the function changes as \(x\) varies, while keeping \(y\) constant. Having this allows for further analysis through mixed partial derivatives by taking these gradients in conjunction with additional variables.
Partial Derivative with Respect to y
The partial derivative with respect to \(y\) is determined in a similar manner. For the function \(z = x^2 \exp(-y^2)\), \(y\) is treated as the variable, while \(x\) behaves as a constant throughout the differentiation process.

Differentiating with respect to \(y\) requires the chain rule due to the \(\exp(-y^2)\) term. This results in \(z_y = -2x^2y \exp(-y^2)\). This derivative highlights how the function behaves as \(y\) changes, with \(x\) remaining constant. Understanding this allows us to study more complex interactions by considering changes in multiple variables at once.
Verification of Equality of Mixed Derivatives
Verification of the equality of mixed derivatives is an important step in multivariable calculus. It involves checking whether the mixed partial derivatives \(z_{xy}\) and \(z_{yx}\) are equal.

In our exercise, after computing both derivatives, we observe that \(z_{xy} = -4xy \exp(-y^2)\) and \(z_{yx} = -4xy \exp(-y^2)\). They are indeed equal, thus confirming the equality of mixed partial derivatives for this particular function.

This equality is anticipated due to Clairaut's theorem, which states that if the function is continuous and has continuous second partial derivatives, the mixed derivatives are the same. This result is a useful tool in the study of multivariable functions, indicating a stable and consistent response of the function to changes across different variables.

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