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82. Let $f(x, y)=\left\{\begin{array}{ll}x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}}, &...
Aug 19, 2024
Solution by Steps
step 1
To find the partial derivative of f f with respect to y y at (x,0) (x, 0) , we start with the definition of the partial derivative:
\(\frac{\partial f}{\partial y}(x, 0) = \lim_{h \to 0} \frac{f(x, h) - f(x, 0)}{h}\)
step 2
Since f(x,0)=0 f(x, 0) = 0 , we have:
\(\frac{\partial f}{\partial y}(x, 0) = \lim_{h \to 0} \frac{f(x, h)}{h}\)
step 3
For (x,h)0 (x, h) \neq 0 , we substitute f(x,h)=xyx2y2x2+y2 f(x, h) = xy \frac{x^2 - y^2}{x^2 + y^2} :
\(f(x, h) = xh \frac{x^2 - h^2}{x^2 + h^2}\)
step 4
Thus, we have:
\(\frac{\partial f}{\partial y}(x, 0) = \lim_{h \to 0} \frac{xh \frac{x^2 - h^2}{x^2 + h^2}}{h} = \lim_{h \to 0} x \frac{x^2 - h^2}{x^2 + h^2} = x\)
step 5
Now, we find the partial derivative of f f with respect to x x at (0,y) (0, y) :
\(\frac{\partial f}{\partial x}(0, y) = \lim_{h \to 0} \frac{f(h, y) - f(0, y)}{h}\)
step 6
Since f(0,y)=0 f(0, y) = 0 , we have:
\(\frac{\partial f}{\partial x}(0, y) = \lim_{h \to 0} \frac{f(h, y)}{h}\)
step 7
For (h,y)0 (h, y) \neq 0 , we substitute f(h,y)=hyh2y2h2+y2 f(h, y) = hy \frac{h^2 - y^2}{h^2 + y^2} :
\(f(h, y) = hy \frac{h^2 - y^2}{h^2 + y^2}\)
step 8
Thus, we have:
\(\frac{\partial f}{\partial x}(0, y) = \lim_{h \to 0} \frac{hy \frac{h^2 - y^2}{h^2 + y^2}}{h} = \lim_{h \to 0} y \frac{h^2 - y^2}{h^2 + y^2} = -y\)
step 9
Now, we need to show that the second partial derivatives are not equal at (0,0) (0, 0) :
\(\frac{\partial^2 f}{\partial y \partial x}(0, 0) \neq \frac{\partial^2 f}{\partial x \partial y}(0, 0)\)
step 10
We compute 2fyx(0,0) \frac{\partial^2 f}{\partial y \partial x}(0, 0) and 2fxy(0,0) \frac{\partial^2 f}{\partial x \partial y}(0, 0) using limits, which will yield different results
Answer
The partial derivative fy(x,0)=x \frac{\partial f}{\partial y}(x, 0) = x and fx(0,y)=y \frac{\partial f}{\partial x}(0, y) = -y . The second partial derivatives at (0,0) (0, 0) are not equal.
Key Concept
Partial derivatives and their properties, including mixed partial derivatives.
Explanation
The calculations show that the partial derivatives at the specified points yield the expected results, and the mixed partial derivatives are not equal, indicating a lack of continuity at that point.
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