The directional derivative of $f: U \to \Bbb{R}^m$ at $p \in U$ in the direction $u$ is the limit, if it exists, $$\nabla_p f(u) = \lim_{t\to 0}\frac{f(p + tu) - f(p)}{t}.$$ (Often one requires that $|u| = 1$.)}

(a) If $f$ is differentiable at $p$, why is it obvious that the directional derivative exists in each direction $u$?}

(b) Show that the function $f : \Bbb{R}^2 \to \Bbb{R}$ defined by $$f(x,y) = \begin{cases} \frac{x^3y}{x^4 + y^2},& (x,y) \neq (0,0)\\ 0,& (x,y) = (0,0) \end{cases}.$$ has $\nabla_{(0,0)}f(u) = 0$ for all $u$ but is not differentiable at $(0, 0)$.

My only question about that problem is about the differentiability of $f$. I cannot see how to prove that $f$ is not differentiable. I tried to show that the partials derivatives is not continuous, show that $\frac{f(x,y)}{|(x,y)|} \not\to (0,0)$ as $(x,y) \to (0,0)$, but nothing worked. I appreciate any hints.

  • $\begingroup$ Do you know the definition of when a function multivariable function is said to be differentiable at a point? $\endgroup$ – Aryaman Maithani Jan 7 '20 at 12:12
  • $\begingroup$ @AryamanMaithani when there is a linear map $(Df)_p$ such that $f(p+v) - f(p) = (Df)_pv + r(v)$ with $\frac{r(v)}{|v|}$ as $|v| \to 0$. $\endgroup$ – Greg Jan 7 '20 at 12:16
  • $\begingroup$ @AryamanMaithani from the definition $f(v) - f(0) = (Df)_pv + r(v)$. Using the hypothesis about the problem, we have $f(v) = r(v)$, from where did my second attempt come from. $\endgroup$ – Greg Jan 7 '20 at 12:20
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    $\begingroup$ Yes, to show that the last limit isn't zero, consider the sequence $(1/n, 1/n^2)$, that is, approach $(0, 0)$ along the path $y = x^2$. You should get $1/2$. $\endgroup$ – Aryaman Maithani Jan 7 '20 at 12:28
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    $\begingroup$ I would like to add that the lack of continuity of partials would not have proven that the function is non-differentiable. For example, consider $x \mapsto x^2\sin\dfrac{1}{x}$ for $x \neq 0$ and $0 \mapsto 0$. This function is indeed differentiable at $0$ even though the derivative is discontinuous. (I haven't actually checked whether the partial is discontinuous or not.) $\endgroup$ – Aryaman Maithani Jan 7 '20 at 12:34

Elaborating from the comments to give a complete answer.

The function $f$ would be differentiable at $(0, 0)$ if there existed $(\alpha, \beta) \in \mathbb{R}^2$ such that $$\lim_{(h, k) \to (0, 0)}\dfrac{\vert f(0+h, 0+k) - f(0, 0) - \alpha h - \beta k\vert}{\sqrt{h^2 + k^2}} = 0.$$

From this definition, it also follows that if such a pair $(\alpha, \beta)$ does exist, then $\alpha = f_x(0, 0)$ and $\beta = f_y(0, 0)$. Computing these derivatives (using the limit-definition) is straightforward and gives us $(\alpha, \beta) = (0, 0)$.

Thus, the function is differentiable at $(0, 0)$ if and only if

$$\lim_{(h, k) \to (0, 0)}\dfrac{\vert f(0+h, 0+k) - f(0, 0)\vert}{\sqrt{h^2 + k^2}} = 0$$ $$ \iff \lim_{(h, k) \to (0, 0)}\dfrac{\vert f(h, k)\vert}{\sqrt{h^2 + k^2}} = 0$$

The above is what you had arrived at.
Now, I show that the above limit can not equal $0$ (in fact, it does not exist but we don't need that).

Consider the sequence $(h_n, k_n) = \left(\frac{1}{n}, \frac{1}{n^2}\right)$. It is clear that $(h_n, k_n) \to (0, 0)$ and $(h_n, k_n) \neq (0, 0)$ for any $n \in \mathbb{N}$.

However, note that $$\dfrac{\vert f(h_n, k_n)\vert}{\sqrt{h_n^2 + k_n^2}} = \dfrac{\frac{1/n^5}{2/n^4}}{\sqrt{1/n^2 + 1/n^4}} = \dfrac{1}{2}\dfrac{1}{\sqrt{1 + 1/n^2}} \to \dfrac{1}{2} \neq 0.$$

Thus, by sequential criterion, we get that limit is not $0$. (The limit isn't $\frac{1}{2}$ though.)

Thus, we have shown that the function is not differentiable.

Also, even if you could show that the partial(s) are discontinuous, you wouldn't have proven the non-differentiability. For example, consider the following function $g:\mathbb{R}^2 \to \mathbb{R}.$

$$g(x, y) = \begin{cases} x^2\sin\left(\dfrac{1}{x}\right),& x \neq 0\\ 0,& x = 0 \end{cases}.$$

It is easy to show that the function is indeed differentiable at $(0, 0)$ using the above definition but the partial derivative with respect to $x$ is not continuous.

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    $\begingroup$ That's a nice answer! Very clear! Thank you. $\endgroup$ – Greg Jan 8 '20 at 10:49

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