Melissa Blanchard and Stuart Graves

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Melissa Blanchard and Stuart Graves

For each of the functions in part (1), what can you say about the relationship between the mixed partials (f_xy(x,y) and f_yx(x,y))?

Part 1 Trials:

For any constant the derivatives will be zero, therefore, trivially f_xy(x,y) = _fyx(x,y).

For x²/2, the partial derivative with respect to y is 0; therefore, f_xy(x,y) = 0. The partial derivative of x²/2 with respect to x is x; therefore, f_yx(x,y) = 0.

For -x² - y⁴+y² (twin peaks) f_x(x,y) = -2x and f_y(x,y) = -4y³+2y. The partial derivative of f_x(x,y) with respect to y is equal to 0. Likewise, the partial derivative of f_y(x,y) with respect to x is also equal to 0.

Part 1 Results:
For each trial from part (1) the values obtained from f_xy(x,y) and f_yx(x,y) were the same.

Part 2:
Now enter the function f(x,y) = 4xy (x² – y²)/(x² + y²) in the second demo. Look at the curves on the first partial derivative graphs which display the second derivatives. Are any of the second partial derivatives defined at (x,y) = (0,0)?

Yes, f_xy(x,y) and f_yx(x,y) are both defined at (x,y) = (0,0). However f_xx(x,y) and f_yy(x,y) are not defined at the origin.

Calculating the partials by hand confirms that f_xy(x,y) and f_yx(x,y) are equal which agrees with the observations from part 1.

f_yx(x,y) = -4(x⁶+9x⁴y² - 9x²y⁴-y⁶) / (x² + y²)³
f_xy(x,y) = -4(x⁶+9x⁴y² - 9x²y⁴-y⁶) / (x² + y²)³

Part 2 Images:

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