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For $$f(x,y)=\begin{cases} \frac{x|y|}{\sqrt{x^2+y^2}} & \text{ for }(x,y)\neq (0,0)\\ 0 & \text{ for } (x,y)=(0,0) \end{cases}$$

I'm trying to prove $f$ is not differentiable at $(0,0)$. I showed if $f$ is differentiable at $(0,0),$ then $A=Df_{(0,0)}=0.$ But I don't know how this lead to a contradiction. Anyone has ideas?

Michael Hardy
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Kelan
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3 Answers3

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If you approach with the paths $(t,0)$ and $(0,t)$, you will conclude that the partial derivatives are $0$.

Zero is particularly convenient because in general, when both partial derivatives are $0$, then any directional derivative must also be $0$ unless $f$ was not differentiable at that point.

Thus in order to show that it is not differentiable at $(0,0)$ it suffices to show a linear path that leads to a different derivative.

One path you may consider is $(t,t)$: For $t \neq 0$, $f(t,t)=\frac{t |{t}|}{\sqrt{2t^2}} =\frac{t {|t|}}{\sqrt{2} {|t|}} =\frac{t}{\sqrt{2}}$ which leads a derivative of $\frac{1}{\sqrt{2}} \neq 0$.

Note: As mentioned in the comments, it should be clear that the argument is only valid because both partial derivatives are zero. In the general non-zero case, you may use a similar argument by defining a second function $g$ as $f$ minus the linear function that makes both partial derivatives of $g$ equal to zero. $f$ is derivable at a point if and only if $g$ is, thus it would suffice to find a problematic path for $g$ to conclude the non-differentiability of $f$.

Carlos Pinzón
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    Obtaining differing directional derivatives along different paths toward the origin should be expected. For example, consider $f(x,y)=x^2+y^3$. The directional derivative of $f$ at $(1,1)$ in the direction of $u_1=(1/\sqrt{2},1/\sqrt{2})$ equals $l_1'(0)$ where $l_1(t)=f(1+t/\sqrt{2},1+t/\sqrt{2})$ which is $5/\sqrt{2}$. If we instead consider the direction $u_2=(1/\sqrt{2},-1/\sqrt{2})$ we get a directional derivative of $l_2'(0)$ where $l_2(t)=f(1+t/\sqrt{2},1-t/\sqrt{2})$ which is $-1/\sqrt{2}$. These differing values does not imply $f$ isn't differentiable at $(1,1)$ – Matthew H. Aug 16 '21 at 00:10
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    But with both partials coming out to $0$ at the origin, then if $f$ were differentiable there, the tangent plane would be horizontal and all the directional derivatives would be $0$, so the path argument is sufficient in this case. – Ned Aug 16 '21 at 00:34
  • I edited the answer because you were both right: It was misleading, but it was correct. It should be clear now. – Carlos Pinzón Aug 16 '21 at 13:33
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It's easy. You only have to see that if $f$ is differentiable in $(0,0)$, then $f'(0,0)$ is a linear transformation. So:

$$f'(0,0)\cdot (1,1)=f'(0,0)\cdot((1,0)+(0,1))=f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$$

And that is a contradiction, because $f'(0,0)\cdot (1,1)\not = f'(0,0)\cdot(1,0)+f'(0,0)\cdot(0,1).$

Manatee
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Prove by contradiction. Suppose the contrary that $f$ is differentiable at $(0,0)$. In particular, partial derivatives $f_{x}(0,0)$ and $f_{y}(0,0)$ exist.

By direct calculation, \begin{eqnarray*} f_{x}(0,0) & = & \lim_{h\rightarrow0}\frac{f(0+h,0)-f(0,0)}{h}\\ & = & 0 \end{eqnarray*} because for $h\neq0$, $f(h,0)=0$. Also, \begin{eqnarray*} f_{y}(0,0) & = & \lim_{h\rightarrow0}\frac{f(0,0+h)-f(0,0)}{h}\\ & = & 0. \end{eqnarray*}

Define the so-called remainder part $R$, $R:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \begin{eqnarray*} R(x,y) & = & \left[f(x,y)-f(0,0)\right]-\left[f_{x}(0,0)x+f_{y}(0,0)y\right]\\ & = & f(x,y). \end{eqnarray*} That $f$ is differentiable at $(0,0)$ implies that $\lim_{(x,y)\rightarrow(0,0)}\frac{1}{\sqrt{x^{2}+y^{2}}}R(x,y)=0.$ In particular, if we put $x=y=t$ and let $t\rightarrow0+$, then $(x,y)\rightarrow(0,0)$ (along the straight line $y=x$). Hence, $\lim_{t\rightarrow0+}\frac{1}{\sqrt{t^{2}+t^{2}}}R(t,t)=\lim_{(x,y)\rightarrow(0,0)}\frac{1}{\sqrt{x^{2}+y^{2}}}R(x,y)=0.$ However, by direct computation, \begin{eqnarray*} \lim_{t\rightarrow0+}\frac{1}{\sqrt{t^{2}+t^{2}}}R(t,t) & = & \lim_{t\rightarrow0+}\frac{1}{\sqrt{2}t}f(t,t)\\ & = & \frac{1}{\sqrt{2}}\lim_{t\rightarrow0+}\frac{1}{t}\cdot\frac{t^{2}}{\sqrt{t^{2}+t^{2}}}\\ & = & \frac{1}{2}, \end{eqnarray*} which is a contradiction.

Danny Pak-Keung Chan
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