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Chapter 12: Problem 42

Verify that \(f_{x y}=f_{y x}\) for the following functions. $$f(x, y)=3 x^{2} y^{-1}-2 x^{-1} y^{2}$$

Short Answer

Expert verified
Question: Verify that \(f_{xy} = f_{yx}\) for the function \(f(x, y) = 3x^2y^{-1} - 2x^{-1}y^2\). Answer: After computing the second partial derivatives, we find that \(f_{xy} = -6x y^{-2} + 4x^{-2}y\) and \(f_{yx} = -6x y^{-2} + 4x^{-2}y\). Since both expressions are equal, we can verify that \(f_{xy} = f_{yx}\) for the given function.

Step by step solution

01

Finding the first partial derivatives \(f_x\) and \(f_y\)

First, we will take the partial derivative of the function with respect to x to find \(f_x\): $$ f_x = \frac{\partial}{\partial x} \left(3x^2y^{-1} - 2x^{-1}y^2\right) $$ Using the power rule, we obtain: $$ f_x = 6x y^{-1} + 2x^{-2}y^2 $$ Now we will find the partial derivative of the function with respect to y to get \(f_y\): $$ f_y = \frac{\partial}{\partial y} \left(3x^2y^{-1} - 2x^{-1}y^2\right) $$ Using the power rule, we obtain: $$ f_y = -3x^2 y^{-2} - 4x^{-1}y $$
02

Finding second partial derivatives \(f_{xy}\) and \(f_{yx}\)

Now we will find the second partial derivatives. Let's start with finding the partial derivative of \(f_x\) with respect to y to find \(f_{xy}\): $$ f_{xy} = \frac{\partial}{\partial y} \left(6x y^{-1} + 2x^{-2}y^2\right) $$ Applying the power rule, we obtain: $$ f_{xy} = -6x y^{-2} + 4x^{-2}y $$ Next, we will find the partial derivative of \(f_y\) with respect to x to find \(f_{yx}\): $$ f_{yx} = \frac{\partial}{\partial x} \left(-3x^2 y^{-2} - 4x^{-1}y\right) $$ Applying the power rule, we obtain: $$ f_{yx} = -6x y^{-2} + 4x^{-2}y $$
03

Comparing \(f_{xy}\) and \(f_{yx}\)

Now, we will compare the expressions for \(f_{xy}\) and \(f_{yx}\): $$ f_{xy} = -6x y^{-2} + 4x^{-2}y $$ $$ f_{yx} = -6x y^{-2} + 4x^{-2}y $$ We can see that \(f_{xy} = f_{yx}\). Therefore, we have verified that \(f_{xy}=f_{yx}\) for the given function \(f(x, y)=3x^2y^{-1}-2x^{-1}y^2\).

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