# Second Order Partial Derivative of functions

I am required to show that all partial order derivatives of the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ defined by

$$f(x,y) = \begin{cases} \dfrac{xy(x^2-y^3)}{x^3+y^3} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y) = (0,0) \end{cases}$$

exists everywhere, but the mixed second order partial derivatives are unequal at the $(0,0).$

I tried to solve it as follows:
find $D_{11}$, $D_{12}$, $D_{21}$ and $D_{22}$ of the function, and thereafter try to show that at $D_{12}(0,0) \neq D_{21}(0,0).$

With some tedious calculations $$D_{12} = \frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3} \\ D_{21} = \frac{9x^4y^2-y^6-9y^4x^2+x^6}{(x^2+y^2)^3}$$

My doubts are:

1. At $(0,0)$ both $D_{12}$ and $D_{21}$ work out to $0$ only so I am going wrong somewhere.

2. I am getting confused with initial condition $f(x,y) = 0$ if $(x,y) = (0,0).$ Because if $(x,y)=(0,0)$ and we need to show inequality at $(0,0)$, isn't the function $0$ and hence both $D_{12}$ and $D_{21}$, therefore $0$, $0$?

3. Is there any good textbook/link on web where I can get more practice problems similar to this?

Request guide to understand the problem

• I did see an almost similar question for first order partial derivative on this forum... but I have still not understood it – SAK Nov 5 '16 at 18:13

First, $f$ is a quotient of polynomials with no singularities except at $(0,0)$, so it is smooth everywhere except at $(0,0)$.

Second, this only makes sense if you use the (more) "real" definition of the derivative, using limits, i.e: $$\partial_x g(x,y)|_{a,b} = \lim_{h\rightarrow 0} (g(a+h,b) - g(a,b))/h$$ Note that your expressions are, literally, identical: $D_{12}=D_{21}$, so you would need some other information to see they are not the same.

We are looking at: $$f(x,y) = \frac{xy(x^2-y^3)}{x^3 + y^3}$$

Here, just use the definition starting from the first derivatives: $$f_x=\partial_x f = \frac{ x^2(3+2x)y^4 - y^7 }{(x^3 + y^3)^2} \;\;\;\;\&\;\;\;\; f_y=\partial_y f = \frac{x^6 - 2x^3(1+2x)y^3 - xy^6}{(x^3 + y^3)^2}$$ Then, we can do: \begin{align} \partial_{x}f_y|_{0,0} &= \lim_{h\rightarrow 0} (f_y(0+h,0) - f_y(0,0))/h \\ &= \lim_{h\rightarrow 0} f_y(h,0)/h \\ &= \lim_{h\rightarrow 0} \frac{h^6}{h^6}\frac{1}{h} \\ &= \infty \end{align} \begin{align} \partial_{y}f_x|_{0,0} &= \lim_{h\rightarrow 0} (f_x(0,0+h) - f_x(0,0))/h \\ &= \lim_{h\rightarrow 0} f_x(0,h)/h \\ &= \lim_{h\rightarrow 0} \frac{ -h^7 }{h^6} \\ &= 0 \end{align} which shows that $\partial_{xy}f(0,0)\ne\partial_{yx}f(0,0)$. Hopefully I didn't make any mistakes!