# How do I show that all directional derivative of $f$ exist at $(0,0)$, but $f$ is not differentiable at $(0,0)$.

For $$(x,y)^2\in \mathbb{R}^2$$, let

$$f(x,y)=\begin{cases} [(2x^2-y)(y-x^2)]^{1/4}&x^2\leq y \leq 2x^2\\ 0& \text{otherwise}\\ \end{cases}$$

show that all directional derivative of $$f$$ exist at $$(0,0)$$, but $$f$$ is not differentiable at $$(0,0)$$.

My attempt: Firstly, I observed that the curve become linear when it approaches to zero.

Let $$u=(u_1,u_2)\in \mathbb{R}^2$$ be a unit vector.

$$D_uf(0,0)=\lim_{t \rightarrow 0}\frac{f(tu_1,tu_2)-f(0,0)}{t}=\lim_{t \rightarrow 0}\frac{0-0}{t}=0.$$

This implies all the directional derivatives of $$f$$ exist at $$(0,0)$$.

I want to improve the more justification, why $$f(tu_1,tu_2)=0$$. I understand with graph of the curve. Can anyone suggest me how I improve my justification in this question.

• Is it possible you miss something in conditions or task itself? – zkutch Jul 7 '20 at 6:27
• Which condition? – User124356 Jul 7 '20 at 6:33
• Hello likely IU student. If you are preparing for Tier 1, feel free to email me for the resources I used for it. – Alfred Yerger Jul 7 '20 at 7:16

Let $$g(t)=(t,\frac32t^2)$$. Then $$g(t)$$ is differentiable at $$t=0$$ and $$g(0)=(0,0)$$. If $$f(x,y)$$ were differentiable at $$(0,0)$$ then its derivative there would be $$0$$ and by the chain rule: $$\frac{{\rm d}f(g(t))}{{\rm d}t\qquad}\left|_{t=0}\right.=\nabla f(0,0)\cdot g'(0)=0$$

However if you work out $$\frac{{\rm d}f(g(t))}{{\rm d}t\quad}\left|_{t=0}\right.$$, it will not be $$0$$.

• This is same as my answer except that you write $1.5$ as $\frac 3 2$. What is the point in repeating an answer. – Kavi Rama Murthy Jul 7 '20 at 11:59
• Its a slightly different perspective - exploiting the differential as a linear map, rather than for its implications for the speed of growth at the origin. I found it appealing to write it this way. – tkf Jul 7 '20 at 12:22

Hint: The inequalities $$t^{2}u^{2}\leq tv \leq 2t^{2} u^{2}$$ will automatically fail for $$|t|$$ sufficiently small and hence $$f(tu,tv)=0$$ for such $$t$$. [I will add more details if you cannot justify this].

Since the partial derivatives vanish at the origin the only candidate for the derivative at that point is $$0$$. So, is $$f$$is differentiable at $$(0,0)$$ then we must have $$\frac {f(x,y)} {\sqrt {x^{2}+y^{2}}} \to 0$$ as $$(x,y) \to 0$$. Get a contradiction by taking $$y=(1.5)x^{2}$$ and letting $$x \to 0$$.

• Yes. I got it. Can you give me some hint for 2nd part. – User124356 Jul 7 '20 at 6:34
• @User124356 I have edited my answer. – Kavi Rama Murthy Jul 7 '20 at 7:07