Answer to Question #249194 in Calculus for moe

Question #249194

True or False:

The mixed partial derivatives "\\displaystyle{\\frac{\\partial^2 z}{\\partial y \\partial x}}"  and "\\displaystyle{\\frac{\\partial^2 z}{\\partial x \\partial y}}"   are equal


1
Expert's answer
2021-10-12T11:06:11-0400

Answer is false.

The mixed partial derivatives "\\frac{\\partial\u00b2Z}{\\partial{x}\\partial{y}} and \\frac{\\partial\u00b2Z}{\\partial{y}\\partial{x}}" are not necessarily equal. If they are continuous at a point then equal at that point. For example let us consider a piece wise defined function,

"Z = \\begin{cases}\n xy\\frac{x\u00b2-y\u00b2}{x\u00b2+y\u00b2} &\\text{if } (x,y)\u2260(0,0) \\\\\n 0 &\\text{if } (x,y)=(0,0) \n\\end{cases}"

=> "Z = \\begin{cases}\n \\frac{x\u00b3y-xy\u00b3}{x\u00b2+y\u00b2} &\\text{if } (x,y)\u2260(0,0) \\\\\n 0 &\\text{if } (x,y)=(0,0) \n\\end{cases}"

Here the symbols "f_{x} , f_{y}, f_{xy}and f_{yx}" are used for partial and mixed partial derivatives


"f_{x}=\\begin{cases}\n \\frac{(x\u00b2+y\u00b2)(y\u00b3-3x\u00b2y)-(xy\u00b3-x\u00b3y)2x}{(x\u00b2+y\u00b2)\u00b2}&\\text{if } (x,y)\u2260(0,0) \\\\\n 0 &\\text{if } (x,y)=(0,0) \n\\end{cases}"

"f_{x}=\\begin{cases}\n \\frac{(3x\u00b2-y\u00b3)}{x\u00b2+y\u00b2}-\\frac{2x(x\u00b3y-xy\u00b3)}{(x\u00b2+y\u00b2)\u00b2})&\\text{if } (x,y)\u2260(0,0) \\\\\n 0 &\\text{if } (x,y)=(0,0) \n\\end{cases}" Similarly

"f_{y}=\\begin{cases}\n \\frac{(x\u00b3-3xy\u00b2)}{x\u00b2+y\u00b2}-\\frac{2y(x\u00b3y-xy\u00b3)}{(x\u00b2+y\u00b2)\u00b2})&\\text{if } (x,y)\u2260(0,0) \\\\\n 0 &\\text{if } (x,y)=(0,0) \n\\end{cases}" So

"f_{xy}(0,0)= lim_{h->0}\\frac{f_{x}(0,h)-f_{x}(0,0)}{h}= lim_{h->0}\\frac{-h\u00b3}{h\u00b2h}=-1"

"f_{yx}(0,0)= lim_{h->0}\\frac{f_{y}(h,0)-f_{y}(0,0)}{h}= lim_{h->0}\\frac{h\u00b3}{h\u00b2h}=1"



"(\\frac{\\partial\u00b2Z}{\\partial{x}\\partial{y}})_{(0,0)} = -1 and (\\frac{\\partial\u00b2Z}{\\partial{y}\\partial{x}} )_{(0,0)} = 1"

So "\\frac{\\partial\u00b2Z}{\\partial{x}\\partial{y}} \u2260\\frac{\\partial\u00b2Z}{\\partial{y}\\partial{x}}"




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