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Determine for the function $$ f: \mathbb{R}^2 \rightarrow \mathbb{R}, \quad f(x, y)= \begin{cases}\frac{x^3 y-x y^3}{x^2+y^2}, & \text { if }(x, y) \neq(0,0), \\ 0, & \text { else, }\end{cases} $$ the partial derivatives $\frac{\partial f}{\partial x}(x, y)$ and $\frac{\partial f}{\partial y}(x, y)$ for $x, y \in \mathbb{R}$ and $\frac{\partial^2 f}{\partial x \partial y}(0,0)$ and $\frac{\partial^2 f}{\partial y \partial x}(0,0)$.

I have determined the first two derivatives $$\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}$$ and $$\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2} $$ But I have no idea how to determine the last two derivatives.

max423
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1 Answers1

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First of all,

$\dfrac{\partial f}{\partial x}(x,y)=\dfrac{y\left(x^4\!+\!4x^2y^2\!-\!y^4\right)}{\left(x^2\!+\!y^2\right)^2}\;\;$ for any $\,(x,y)\neq(0,0)\;,$

$\dfrac{\partial f}{\partial y}(x,y)=\dfrac{x\left(x^4\!-\!4x^2y^2\!-\!y^4\right)}{\left(x^2\!+\!y^2\right)^2}\;\;$ for any $\,(x,y)\neq(0,0)\;.$

Moreover ,

$\dfrac{\partial f}{\partial x}(0,0)=\lim\limits_{x\to0}\dfrac{f(x,0)-f(0,0)}x=\lim\limits_{x\to0}\dfrac0x=0\;\;,$

$\dfrac{\partial f}{\partial y}(0,0)=\lim\limits_{y\to0}\dfrac{f(0,y)-f(0,0)}y=\lim\limits_{y\to0}\dfrac0y=0\;\;,$

$\dfrac{\partial^2 f}{\partial x\partial y}(0,0)=\lim\limits_{x\to0}\dfrac{\frac{\partial}{\partial y}f(x,0)-\frac{\partial}{\partial y}f(0,0)}x=\lim\limits_{x\to0}\dfrac xx=1\;\;,$

$\dfrac{\partial^2 f}{\partial y\partial x}(0,0)=\lim\limits_{y\to0}\dfrac{\frac{\partial}{\partial x}f(0,y)-\frac{\partial}{\partial x}f(0,0)}y=\lim\limits_{y\to0}\dfrac{-y}y=-1\;.$

Angelo
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