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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 $(0,0)$ ?

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    $\begingroup$ Can you find the derivative of $x\mapsto |x|$ where it exists? $\endgroup$
    – Git Gud
    Dec 29 '18 at 21:05
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    $\begingroup$ 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? $\endgroup$ Dec 29 '18 at 21:08
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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 $x\to 0$ $$ \rho(x,x^2)=\frac{|x|^{1/2}x^4}{2x^4\cdot |x|\sqrt{1+x^2}}=\frac{1}{2|x|^{1/2}\sqrt{1+x^2}}\not\to 0. $$

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

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  • $\begingroup$ @A.Γ. No, in the problem I have it is $x^4 + y^2$. Is it a mistake in the problem? $\endgroup$ Dec 29 '18 at 21:49
  • $\begingroup$ I have made a mistake in the problem, the denominator is $ x^4 + y^2$. $\endgroup$ Dec 29 '18 at 22:04
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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 $(0,0)$.

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  • $\begingroup$ Using the same argment can we say that $f $ is not continuous on $(0,0)$ ? $\endgroup$ Dec 29 '18 at 22:43
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    $\begingroup$ @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 $(x,y)$. $\endgroup$ Dec 29 '18 at 22:45
  • $\begingroup$ Thank you so much for your help. $\endgroup$ Dec 29 '18 at 23:03

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