Let $$f(x,y) = \begin{cases} \dfrac{x^3y}{x^2+y^2} & \text{if $(x,y)\neq0$} \\ 0 & \text{if $(x,y)=0$} \end{cases}$$
I've to show that :
$$\partial^2_{xy}(0,0)=1\neq\partial^2_{yx}(0,0)=0$$ and also show that $\partial^2_{xy}(x,y)$ is not continuous at $0$ ?
I'm not getting the correct answer for $\partial^2_{xy}f(0,0)$ and $\partial^2_{yx}f(0,0)$..also don't know how to check $\partial^2_{xy}(x,y)$ is not continuous.
can anyone just explain this to me..thanks in advance
On
As to $\partial^2_{yx}(0,0)$, we use the standard definition of the derivative. We first calculate $\frac{\partial}{\partial x} f (0,0)$. We know: $$\frac{\partial}{\partial x} f(x_0,y_0)=\lim_{t\rightarrow 0}\frac{f(x_0+t,y_0)-f(x_0,y_0)}{t}$$ So in this case: $$=\lim_{t\rightarrow 0}\frac{f(0+t,0)-f(0,0)}{t}=\lim_{t\rightarrow0}\frac{f(t,0)}{t}=\lim_{t\rightarrow 0}\frac{\left(\frac{0}{t^2}\right)}{t}=\lim_{t\rightarrow 0}0=0$$ So now, $$\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x} f\right)(0,0)=\frac{\partial}{\partial y} 0=0$$
As for the continuity, it is not continuous. We know the derivative $\partial^2_{xy}$ is equal to $$\frac{x^6+6x^4y^2-3x^2y^4}{(x^2+y^2)^3}$$ We now take two limits, $\lim_{x\rightarrow 0}$ and $\lim_{y\rightarrow 0}$. The limit for $x$ is equal to $$\frac{0+0+0}{y^6}=0$$ and the limit for $y$ is $$\frac{x^6+0-0}{x^6}=1$$
So, the limit is path-dependent, so the $\partial^2_{xy}$ is not continuous!
$$\lim_{x\rightarrow 0} \partial_xf(x,y)=0$$ $$\lim_{y\rightarrow 0}\partial_yf(x,y)=x$$ so $\partial^2_{xy}f(0,0)=0$ while $\partial^2_{yx}f(0,0)=1$.
By the contraposition of Schwarz theorem, this proves that the second derivatives of $f$ are not continuous in $(0,0)$.
Another proof that $\partial^2_{xy} f$ is not continuous in $(0,0)$, without invoking Schwarz theorem, is that $\partial^2_{xy}f(x,y)$ doesn't have a limit in $(0,0)$.