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Define $F:\mathbb{R}^2\to\mathbb{R}$ by $F(x,y)=y^2+\sqrt{y^2x^4}+x$.

Determine if F is differentiable at (1,0) or not.

To solve this problem, I have tried to applied $\lim_{x \to a} \frac{||F(x)-F(a)-T(x-a)||}{||x-a||}$ to figure out whether the limit is equal to zero or not.

Then I got $\lim_{(x,y) \to (1,0)}\frac{y^2+yx^2-y}{\sqrt{(x-1)^2+y^2}}$ , the problem is I do not know how to solve such a limit or I used a wrong method to determine the differentiability.

In the other sub-part of the question, I have already solved the function is differentiable at (1,1) and (0,0).

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    $\begingroup$ $F$ is not a vector valued function. What is $T$? $\endgroup$ – uniquesolution Mar 6 at 18:19
  • $\begingroup$ I thought the existence of partial derivatives at a point were a necessary condition for differentiability at that point. At $(0,0)$ they don't exist do they? so how did you prove it is differentiable at $(0,0)$? They don't seem to exist at $(1,0)$ either. Don't take my word, I'm also a student so hopefully someone will answer this soon (and as mentioned above, F is not a vector valued function since it takes values in reals - it is real valued.) $\endgroup$ – Displayname Mar 6 at 18:34
  • $\begingroup$ @Displayname The partials do exist at $(0,0).$ $\endgroup$ – zhw. Mar 6 at 18:37
  • $\begingroup$ @zhw im a bit confused now...so what would the partial of $F$ w.r.t y be at (0,0)? $\endgroup$ – Displayname Mar 6 at 18:42
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    $\begingroup$ @Displayname $F(0,y)=y^2,$ so the partial of $F$ wrt $y$ is $0$ at the origin. $\endgroup$ – zhw. Mar 6 at 19:01
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Hint: $F(x,y)=y^2+ |y|x^2 +x.$ Does $\partial F/\partial y (1,0)$ exist?

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  • $\begingroup$ I see, as one of the partial derivative does not exist at (1,0), therefore F is not differentiable at (1,0). Thanks a lot for your help. $\endgroup$ – Antony Mar 7 at 2:08

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