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I'm having problems with the following demonstration: let $f:\mathbb{R}^2\rightarrow\mathbb{R} $ a continuous function on $(0,0)$ and let $g:= f(x,y) ||(x,y)||^4$. Prove that $g$ is differentiable on $(0,0)$.

We know by definition that the function $g$ will be differentiable on $(0,0)$ if and only if:

$\lim_{(x,y)\to(0,0)}=$$\frac{f(x,y)- <\nabla g,(x,y)> - g(0,0)}{||(x,y)||}=0$

I started by finding the $\nabla g=(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y})$ to could be able to construct my limit of differentiability and after prove that is zero. But, the problem is we don't know too much about $f$. Notice that:

$\frac{\partial g}{\partial x}(0,0) = \lim_{(t)\to(0)} \frac{g((0,0)+t(1,0)) -g(0,0)}{t}$

$\frac{\partial g}{\partial y}(0,0) = \lim_{(t)\to(0)} \frac{g((0,0)+t(0,1)) -g(0,0)}{t}$

Where we can easily see observing the definition of $g$ that $g(0,0)=0$.

And we also know that $f(x,y)= \frac{g(x,y)}{||(x,y)||^4}$

Having said this how can I find the partial derivatives from what I know? Any hint?

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    $\begingroup$ I think you are mixing some things. Why does $f(0,0)$ not exist? (It does.) Calculate $g_x$ and $g_y$ by definition. Then you can calculate the first limit of your post. $\endgroup$
    – user339727
    Jul 21 '17 at 16:40
  • $\begingroup$ You are right, otherwise won't be continuous on $(0,0)$ $\endgroup$ Jul 21 '17 at 16:45
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The partial derivatives are (using that $f$ is continuos at $(0,0)$), $$ \frac{\partial g}{\partial x}(0,0)=\lim_{t\to 0}\frac{g(t,0)-g(0,0)}{t}=\lim_{t \to 0}f(t,0)t^3=0$$ and $$ \frac{\partial g}{\partial y}(0,0)=\lim_{t\to 0}\frac{g(0,t)-g(0,0)}{t}=\lim_{t \to 0}f(0,t)t^3=0.$$ To prove that $g$ is differentiable we need to prove that $$ \lim_{(x,y) \to (0,0)}\frac{g(x,y)-g(0,0)-x\frac{\partial g}{\partial x}(0,0)-y\frac{\partial g}{\partial y}(0,0)}{\| (x,y)\|}=0.$$ And, indeed, \begin{equation} \begin{aligned} &\lim_{(x,y) \to (0,0)}\frac{g(x,y)-g(0,0)-x\frac{\partial g}{\partial x}(0,0)-y\frac{\partial g}{\partial y}(0,0)}{\| (x,y)\|} \\ &=\lim_{(x,y) \to (0,0)}\frac{g(x,y)}{\| (x,y)\|} \\ &=\lim_{(x,y) \to (0,0)}f(x,y)\|(x,y)\|^3 \\ &=0. \end{aligned} \end{equation}

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  • $\begingroup$ Why did you wrote in the limit of the partial derivatives $f(0,0)$? $\endgroup$ Jul 21 '17 at 16:48
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    $\begingroup$ $g(0,0)=0,$ so first partial derivative is $\lim_{t \to 0}g(t,0)/t=\lim_{t\to 0}f(t,0)t^3.$ Since $f$ is continuous at $(0,0)$ then $f(t,0) \to f(0,0)$ as $t\to 0.$ So $\lim_{t\to 0}f(t,0)t^3=f(0,0) \cdot 0^3.$ Similarly the other. $\endgroup$ Jul 21 '17 at 16:51

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