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Suppose $f:\mathbb{R}^2\to\mathbb{R}$ defined by $f(x,y) = \frac{x^2y}{x^2+y^2}$, such that $x^2+y^2>0$ and $f(0,0) = 0$. Show that for every path $\lambda:(-\epsilon,\epsilon)\to\mathbb{R}^2$ differentiable at $0$, with $\lambda(0)=(0,0)$, $(f\circ \lambda)'(0)$ exists

For me, this question was just about using the chain rule:

$$(f\circ \lambda)'(0) = f'(\lambda(0))\lambda'(0)$$

Since $\lambda(0)=0$ exists and also $\lambda'(0)$, I should just show that $f$ is differentiable.

I'd first take the partials at $x$ and $y$:

$$\frac{\partial f}{\partial x} = \frac{2xy}{x^2+y^2} - \frac{x^2y}{(x^2+y^2)^2}$$

$$\frac{\partial f}{\partial y} = \frac{x^2}{x^2+y^2} - \frac{x^2y}{(x^2+y^2)^2}$$

Now remember that a function $f:\mathbb{R}^2\to\mathbb{R}$ is differentiable if there exists $r(v)$ such that

$$f(a+v) = f(a) + \frac{\partial f}{\partial x}v_1 + \frac{\partial f}{\partial y}v_2 + r(v)$$

where $\frac{r(v)}{|v|}\to 0$

So solving for $r(v)$:

$$r(v) = f(a+v)-f(a) - \frac{\partial f}{\partial x}(a)v_1 - \frac{\partial f}{\partial y}(a)v_2$$

*where $v = (v_1,v_2)$

So:

$$\frac{r(v)}{|v|} = \frac{f(a+v)-f(a) - \frac{\partial f}{\partial x}(a)v_1 - \frac{\partial f}{\partial y}(a)v_2}{\sqrt{v_1^2+v_2^2}} $$

But the partial derivatives are not defined on $0$, what should I do here?

If the limit of that thing goes to $0$ when $v\to 0$, $f$ is differentiable.

UPDATE:

By the comment below, the partials at $0$ are $0$, so we end with the following limit:

$$\lim_{v\to 0}\frac{r(v)}{|v|} = \frac{f(a+v)-f(a)}{\sqrt{v_1^2+v_2^2}} =$$

$$\lim_{(v_1,v_2)\to (0,0)} \frac{v_1^2v_2}{\sqrt{v_1^2+v_2^2}}$$

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  • $\begingroup$ For a function $F:U\subset\mathbb R^2\to\mathbb R$ the partial derivatives are defined as $$\frac{\partial F}{\partial x}(a,b)=\lim_{h\to 0}\frac{F(a+h,b)-F(a,b)}{h}\qquad\text{and}\qquad\frac{\partial F}{\partial y}(a,b)=\lim_{k\to 0}\frac{F(a,b+k)-F(a,b)}{k}$$ whenever those limits exist. Then, for the given function we have \begin{align*} \frac{\partial f}{\partial x}(0,0)&=\lim_{h\to 0}\frac{\frac{h^2\cdot 0}{h^2+0^2}-0}{h}=\lim_{h\to 0}\frac{0}{h}=0\\ \frac{\partial f}{\partial y}(0,0)&=\lim_{k\to 0}\frac{\frac{0^2\cdot k}{0^2+k^2}-0}{k}=\lim_{h\to 0}\frac{0}{k}=0 \end{align*} $\endgroup$ – Ángel Mario Gallegos Apr 14 '17 at 22:43
  • $\begingroup$ @ÁngelMarioGallegos thank you! I updated the question with your observation. $\endgroup$ – Guerlando OCs Apr 14 '17 at 23:15
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Set $\lambda(t) = (\mu(t),\nu(t))$. For now, assume one of $\mu'(0)$ and $\nu'(0)$ is nonzero.

\begin{align*} (f\circ \lambda)'(0) &=\lim\limits_{h\to 0} \frac{f(\lambda(h))-f(\lambda(0))}{h}\\ &= \lim\limits_{h\to 0}\frac{\mu^2(h)\nu(h)}{h(\mu^2(h)+\nu^2(h))}\\ &=\lim\limits_{h\to 0}\frac{\mu(h)}{h}\frac{\mu(h)\nu(h)}{\mu^2(h)+\nu^2(h)}\\ &=\lim\limits_{h\to 0}\frac{\mu(h)}{h}\frac{\left(\frac{\mu(h)}{h}\right)\left(\frac{\nu(h)}{h}\right)}{\left(\frac{\mu(h)}{h}\right)^2+\left(\frac{\nu(h)}{h}\right)^2} =\mu'(0)\frac{\mu'(0)\nu'(0)}{\mu'^2(0)+\nu'^2(0)}. \end{align*} If $\lambda'(0) = 0$, then $\lim\limits_{h\to 0}\frac{\mu(h)}{h} = 0$ and $\left|\frac{\mu(h)\nu(h)}{\mu^2(h)+\nu^2(h)}\right|\leq\frac12$ is bounded, so $(f\circ\lambda)'(0)=0$. In any case, $(f\circ\lambda)'(0)$ always exists.

If $f$ were differentiable, then applying the chain rule $(f\circ\lambda)'(0) = f'(\lambda(0))\lambda'(0) = 0$ for every $\lambda$. However, if we take $\lambda(h) = (h,h)$, we have $(f\circ\lambda)'(0)=1/2$. This means that $f$ is not differentiable.

The point of this exercise is to show you that even if $f:\mathbb R^2\to\mathbb R$ is differentiable in every direction (the direction if $\lambda'(0)$ for every $\lambda$, in this case) it might not be differentiable in the linear map sense. You cannot use the chain rule if you don't know $f$ is differentiable to begin with.

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