Is $f(x,y) = \frac{|x|^{5/2} y}{x^4 + y^2}$ differentiable at $(0,0)$?

Let $$f: \mathbb{R}^2 \rightarrow \mathbb{R}$$ be a function defined as :

$$\begin{cases} f(x,y) = \frac{|x|^{5/2} y}{x^4 + y^2} , & (x,y) \not= (0,0) \\ f(0,0) = 0 \end{cases}$$

1. Compute $$\frac{df}{dx}(0,0)$$ and $$\frac{df}{dy}(0,0)$$.
2. Is $$f$$ differentiable at $$(0,0)$$?

I have problem with question 2.

For $$f$$ to be differentiable, the partial derivatives must exist and be continuous.

They exist because we have:

$$\frac{df}{dx}(0,0) = \lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x} = 0$$

$$\frac{df}{dy}(0,0) = \lim_{y \to 0} \frac{f(0,y) - f(0,0)}{y} = 0$$

but I cannot compute the partial derivatives because of the absolute value.

How to know if the partial derivatives are continuous at $$(0,0)$$ ?

• Can you find the derivative of $x\mapsto |x|$ where it exists? Dec 29 '18 at 21:05
• I need to split the interval to have two differents cases. In this case I have the power $5/2$ which is the problem for me. Why my question is voted to be closed? Dec 29 '18 at 21:08

3 Answers

With your last edit the answer to 2) is negative. For differentiability we need by definition that $$\rho(x,y)=\frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{\sqrt{x^2+y^2}}\to 0$$ as $$(x,y)\to(0,0)$$, however, with $$y=x^2$$ we get for $$x\to 0$$ $$\rho(x,x^2)=\frac{|x|^{1/2}x^4}{2x^4\cdot |x|\sqrt{1+x^2}}=\frac{1}{2|x|^{1/2}\sqrt{1+x^2}}\not\to 0.$$

• Why did divide by $\sqrt{x^2 + y^2}$ ? I am not familiar with the function you considered. Dec 29 '18 at 22:40
• @ZouhairElYaagoubi What is your definition of a differentiable function? I am using the standard one. Here $|h|=|(x,y)|=\sqrt{x^2+y^2}$.
– A.Γ.
Dec 29 '18 at 22:45
• Ah! I see it now. Thank you so much for your help. Accepted answer. Dec 29 '18 at 23:04

$$f$$ is differentiable at $$(0,0)$$ and the derivative is the always vanishing map as for $$(x,y)\neq(0,0)$$ you have

$$0 \le \left\vert \frac {f(x,y)}{\sqrt{x^2+y^2}} \right\vert = \frac{\vert x \vert}{\sqrt{x^2+y^2}} \frac{\vert xy \vert}{x^2+y^2}\sqrt{\vert x \vert}\le \frac{\sqrt{\vert x \vert}}{2} \to 0$$

As $$(x,y) \to (0,0)$$.

Take care! There is indeed a theorem stating that the derivative exists and is continuous if and only if the partial derivatives exist and are continuous. But a map maybe be differentiable even if the partial derivatives exist but are not continuous.

• @A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem? Dec 29 '18 at 21:49
• I have made a mistake in the problem, the denominator is $x^4 + y^2$. Dec 29 '18 at 22:04

To calculate the partial derivatives, recall that the derivative of the absolute value is $$|\cdot|' = \text{sign}$$, on $$\mathbb{R}\setminus \{0\}$$.

Therefore for $$(x,y) \ne (0,0)$$ we have

$$\frac{\partial f}{\partial x}(x,y) = \frac{\frac{5}2(\operatorname{sign} x)|x|^{3/2}y(x^4+y^2) - |x|^{5/2}y\cdot 4x^3}{(x^4+y^2)^2}$$

$$\frac{\partial f}{\partial y}(x,y) = \frac{|x|^{5/2}(x^4+y^2) - |x|^{5/2}y\cdot 2y}{(x^4+y^2)^2}$$

E.g. for $$x > 0$$ and $$y = x^2$$ we get

$$\frac{\partial f}{\partial x}(x,x^2) = \frac{5|x|^{7.5} - 4|x|^{7.5}}{4x^8} = \frac1{4\sqrt{|x|}} \not\to 0$$

as $$(x,y) \to (0,0)$$ so $$\frac{\partial f}{\partial x}$$ isn't continuous at $$(0,0)$$.

This says nothing about differentiability, but the other answers show that $$f$$ is not differentiable at $$(0,0)$$.

• Using the same argment can we say that $f$ is not continuous on $(0,0)$ ? Dec 29 '18 at 22:43
• @ZouhairElYaagoubi $f$ is continuous at $(0,0)$: $$|f(x,y)| = \frac{|x|^{5/2}|y|}{x^4+y^2} \le \frac{r^{5/2}r}{r^2} = r^{3/2} \xrightarrow{r\to 0} 0$$ where $r = \sqrt{x^2+y^2}$ is the norm of $(x,y)$. Dec 29 '18 at 22:45
• Thank you so much for your help. Dec 29 '18 at 23:03