- #1

mathmari

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MHB

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Let's consider the function \begin{align*}f:\mathbb{R}^2&\rightarrow \mathbb{R} \\ (x,y)&\mapsto \sqrt{|x|\cdot |y|}\end{align*}

Show that $f$ is partially differentiable in $(0,0)$ but not total differentiable.I have done the following:

We prove that $f$ is partially differentiable in each direction:

Let $0 \neq v =(r, s) \in \mathbb{R}^2$ be the direction and $0 \neq h \in \mathbb{R}$ :

\begin{align*}\frac{1}{h}\cdot \left (f\left ((0,0)+hv\right )-f(0,0)\right )&=\frac{1}{h}\cdot \left (f(hr,hs)-0\right )=\frac{1}{h}\cdot f(hr,hs)=\frac{1}{h}\cdot \sqrt{|hr|\cdot |hs|}=\frac{1}{h}\cdot \sqrt{|h|^2\cdot |r|\cdot |s|}\\ &=\frac{1}{h}\cdot |h|\cdot \sqrt{|r|\cdot |s|} =\begin{cases}\frac{1}{h}\cdot h\cdot \sqrt{|r|\cdot |s|} & \text{ if } h\geq 0\\ \frac{1}{h}\cdot \left (-h\right )\cdot \sqrt{|r|\cdot |s|} & \text{ if } h< 0\end{cases}\\ &=\begin{cases}\sqrt{|r|\cdot |s|} & \text{ if } h\geq 0\\ - \sqrt{|r|\cdot |s|} & \text{ if } h< 0\end{cases}\end{align*}

Since this term is independent from $h$, the limit for $h \rightarrow 0$ is equal to $\pm \sqrt{|r|\cdot |s|}$,which means that the limit exists.

That means that $f$ is partially differentiable in every direction, since for each $v$ there is a directional derivative.

Is that correct? Or do we show that $f$ partially differentiable by calculating the partial derivatives $f_x$ and $f_y$ ? :unsure:

To show that that $f$ is not totally differentiable do we show that that $f$ is not continuous in $(0,0)$ ? :unsure: