Prove that the function f(x,y) is not differentiable

Question

I have the following problem:

Prove that the function:

$f(x,y)= \ \begin{cases} \frac{x^3-x\cdot y^2}{x^2+y^2} & (x,y)\neq (0,0) \\ \\0 & (x,y)=(0,0) \end{cases} \\$

is continuous on $R^2$ and has its first order partial derivatives. everywhere on $R^2$, but $f$ is not differentiable at $(0,0)$

I know how to prove that it is continuous on $R^2$ and its partial derivatives exist at $(0,0)$ (I use limit definition of a derivative). But I do not know how to prove that this function is not differentiable.

Answer 1

We have $${ f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } }{ x } } =1\\ \\ { f }_{ y }^{ \prime }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac { f\left( 0,y \right) -f\left( 0,0 \right) }{ y } } =\lim _{ y\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } }{ y } } =0\\ f\left( x,y \right) -f\left( 0,0 \right) =\frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } =x+\left( \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x \right) ={ f }_{ x }^{ \prime }\left( 0,0 \right) x+{ f }_{ y }^{ \prime }\left( 0,0 \right) y+\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$ where $\alpha \left( x,y \right) =\frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } } =\frac { -2x{ y }^{ 2 } }{ \left( x^{ 2 }+y^{ 2 } \right) \sqrt { x^{ 2 }+y^{ 2 } } } $ however when $n\rightarrow \infty $ $$\alpha \left( \frac { 1 }{ n } ,\frac { 1 }{ n } \right) =\frac { -\frac { 1 }{ { n }^{ 3 } } }{ \frac { 1 }{ { n }^{ 3 } } \sqrt { 2 } } =-\frac { 1 }{ \sqrt { 2 } } $$ which shows $$\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \neq o\left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \right) $$

Answer 2

At a point of differentiablity of $f$, we have that, for any vector $v=(a,b)$, the derivative of $f$ in the direction of $v$ satisfies $$\frac{\partial f}{\partial v} = a \frac{\partial f}{\partial x} + b \frac{\partial f }{\partial y}$$ However, because $$f(x,y) = \begin{cases} x & \text{ if } x\neq 0, y=0 \\ 0 & \text{ if } x=0 ,y \neq 0 \\ 0 & \text{ of } x=y \neq 0 \\ \end{cases}$$ we can conclude $$ \begin{cases} \frac{\partial f}{\partial x}(0,0) = 1 \\ \frac{\partial f}{\partial y}(0,0) = 0\\ \frac{\partial f}{\partial v}(0,0) = 0 && \text{ when } v=(1,1)\\ \end{cases}$$ and so $f$ must not be differentiable at $(0,0)$.

Answer 3

Hint: Use the definition. Let $\rho(h,k)$ satisfy $\rho (0,0) = 0$ and $$f(h,k) - f(0,0)= f_1(0,0)h + f_2(0,0)k + \rho (h,k)\sqrt{h^2 + k^2}$$ Since you calculated the partial derivatives $f_1$ and $f_2$ at $(0,0)$ you get $$\frac{h^3 - hk^2}{h^2 + k^2} = h + \rho(h,k) \sqrt{h^2 + k^2}$$ Thus we find $$\rho(h,k) = \frac{-2hk^2}{(h^2 + k^2)^{3/2}}$$

You need to show that $\displaystyle\lim_{(h,k) \to (0,0)} \rho (h,k) \neq 0$. Check the path ($h = k$).

Answer 4

Depending on how far you have gone and what definition you are using, but you could for example show that there does not exist a linear mapping that satisfies $f(0+h)=f(0)+L(h)+||h||\epsilon (h)$, where $\epsilon$ term approaches zero as $h$ approaches zero.

Or you could show that partial derivatives are not continuous at origo if you have gone through continuously differentiablity.