# $f(x,y) = \frac{x^3y}{x^4 + y^2}$ is not differentiable at $(0,0)$.

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.

• Do you know the definition of when a function multivariable function is said to be differentiable at a point? – Aryaman Maithani Jan 7 '20 at 12:12
• @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 $|v| \to 0$. – Greg Jan 7 '20 at 12:16
• @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. – Greg Jan 7 '20 at 12:20
• 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 $1/2$. – Aryaman Maithani Jan 7 '20 at 12:28
• 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.) – Aryaman Maithani Jan 7 '20 at 12:34

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.

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