Showing a function $f:\mathbb R^2\to \mathbb R^2$ is partially and totally differentiable

Question

Consider $f:\mathbb R^2 \to \mathbb R, \quad (x,y) \mapsto \frac{x^3}{\sqrt{x^2+y^2}}$ and $(0,0)\mapsto 0$. How do I show that f is totally differentiable at $(0,0)$?

What about showing that a function is partially differentiable? Consider a similar function $f(x,y)=\frac{x^2y}{x^4+y^2}$ and $f(0,0)=0.$ I have shown that this is not continuous at $0$, since I can approach $(0,0)$ using different functions for $y$ and the result of the limit will be different. How do I show that partial derivatives exist? Is merely showing that the normal derivative limit (but with $x$ or $y$ fixed) exists?

Answer 1

Partial derivatives in $f^{'}_{x}(0,0)=f^{'}_{y}(0,0)=0$. So for differentiable we have $$f(\Delta x, \Delta y) = \epsilon(\Delta x, \Delta y) \sqrt{x^2+y^2}$$ so $$\epsilon(\Delta x, \Delta y) = \frac{\Delta x^3}{\Delta x^2+\Delta y^2} \leqslant \Delta x \rightarrow 0$$

Answer 2

Note that$$\lim_{(x,y)\to(0,0)}\frac{x^3}{x^2+y^2}=\lim_{(x,y)\to(0,0)}x\overbrace{\frac{x^2}{x^2+y^2}}^{\phantom1\leqslant1}=0.$$But, if $L\colon\Bbb R^2\longrightarrow\Bbb R$ is the null function, then $L$ is linear and$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-L(x,y)}{\|(x,y)\|}=0.$$So, by definition, the total derivative of $f$ at $(0,0)$ is $L$. And therefore $\frac{\partial f}{\partial x}(0,0)=\frac{\partial f}{\partial y}(0,0)=0$.

Answer 3

You can prove in this way: $$x\leq\sqrt{x^2+y^2}$$ $$\frac {x}{\sqrt{x^2+y^2}}\leq1$$ $$x^2\times\frac{x}{\sqrt{x^2+y^2}}\leq x^2\times 1$$ $$\lim_{(x,y)\to(0,0)} \frac {x^3}{\sqrt{x^2+y^2}}\leq \lim_{(x,y)\to(0,0)} x^2=0$$