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Chapter 13: Problem 38

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 the mixed second-order partial derivatives of the function \(f(x, y) = 3x^2 y^{-1} - 2x^{-1} y^2\) are equal. Solution: After calculating the second-order partial derivatives, we obtained \(f_{x y} = (-6x)(y^{-2}) - (4)(x^{-2})(y)\) and \(f_{y x} = (6x)(y^{-2}) - (-4)(x^{-2})(y)\). Since \(f_{x y} = f_{y x}\), we can verify that the mixed second-order partial derivatives are equal.

Step by step solution

01

Calculate \(f_x\) (First Derivative with respect to x)

Differentiate the function f(x, y) with respect to x: $$f_x = \frac{\partial}{\partial x}[3x^2 y^{-1} - 2x^{-1} y^2]$$ Using the power rule of differentiation, we get: $$f_x = (6x) (y^{-1}) - (-2)(x^{-2})(y^2)$$
02

Calculate \(f_{x y}\) (Second Derivative with respect to y after x)

Next, differentiate \(f_x\) with respect to y: $$f_{x y} = \frac{\partial}{\partial y}[(6x) (y^{-1}) - (-2)(x^{-2})(y^2)]$$ Using the power rule of differentiation again, we obtain: $$f_{x y} = (-6x)(y^{-2}) - (4)(x^{-2})(y)$$
03

Calculate \(f_y\) (First Derivative with respect to y)

Now, differentiate the original function f(x, y) with respect to y: $$f_y = \frac{\partial}{\partial y}[3x^2 y^{-1} - 2x^{-1} y^2]$$ Using the power rule, we get: $$f_y = (-3)(x^2)(y^{-2}) - (4)(x^{-1})(y)$$
04

Calculate \(f_{y x}\) (Second Derivative with respect to x after y)

Lastly, differentiate \(f_y\) with respect to x: $$f_{y x} = \frac{\partial}{\partial x}[(-3)(x^2)(y^{-2}) - (4)(x^{-1})(y)]$$ Using the power rule of differentiation, we obtain: $$f_{y x} = (6x)(y^{-2}) - (-4)(x^{-2})(y)$$
05

Compare \(f_{x y}\) and \(f_{y x}\)

Now, compare the two mixed second-order partial derivatives: $$f_{x y} = (-6x)(y^{-2}) - (4)(x^{-2})(y)$$ $$f_{y x} = (6x)(y^{-2}) - (-4)(x^{-2})(y)$$ As you can see, \(f_{x y}\) and \(f_{y x}\) are equal. Therefore, we can verify that: $$f_{x y} = f_{y x}$$

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