Find if $$ \begin{cases} \frac{x^3+y^4}{x^2+y^2} & (x,y,z)\neq (0,0,0)\\ 0 & (x,y,z)=(0,0,0) \end{cases} $$ is differentiable
So what I am using is the facts:
- If the partial derivatives are continuous, then it is differentiable.
- Differentiability by definition
Starting with checking that the function is continuous at $(0,0)$
$$ \lim_{(x,y)\to (0,0)}\frac{x^3+y^4}{x^2+y^2}\leq \lim_{(x,y)\to (0,0)}\left|\frac{x^3}{x^2+y^2}\right|+\left|\frac{y^4}{x^2+y^2}\right|\leq \lim_{(x,y)\to (0,0)}|x|+|y^2|=0 $$
The partial derivatives at $(x,y)\neq (0,0)$ are:
$$ f_x=\frac{3x^2(x^2+y^2)-2x(x^3+y^4)}{(x^2+y^2)^2}=\frac{x^4+3x^2y^2-2xy^4}{(x^2+y^2)^2} $$
and
$$ f_y=\frac{4y^3(x^2+y^2)-2y(x^3+y^4)}{(x^2+y^2)^2}=\frac{4x^2y^3+2y^5-2x^3y}{(x^2+y^2)^2} $$
The partial derivatives at $(x,y)=(0,0)$ are:
$$ f_x(0,0)=\lim_{\Delta x\to 0}\frac{\frac{(\Delta x)^3}{(\Delta x)^2}}{\Delta x}=1 $$
$$ f_y(0,0)=\lim_{\Delta y\to 0}\frac{\frac{(\Delta y)^4}{(\Delta y)^2}}{\Delta y}=\lim_{\Delta y\to 0} \Delta y=0 $$
It seems like that the partial derivatives are not continuous (so it will be easier to prove by definition):
$$ \lim_{(\Delta x, \Delta y)\to (0,0)} \frac{\frac{(\Delta x)^3+(\Delta y)^4}{(\Delta x)^2+(\Delta y)^2}-\Delta x}{\sqrt{(\Delta x)^2+(\Delta y)^2}} $$
Using polar coordinates:
$$ \lim_{r\to 0} \frac{\frac{r^3\cos^3 \theta+r^4\sin^4 \theta}{r^2}-r\cos \theta}{r}=\lim_{r\to 0} \frac{{r\cos^3 \theta+r^2\sin^4 \theta}-r\cos \theta}{r}=\\=\lim_{r\to 0} \cos^3\theta +r\sin^4 \theta -\cos \theta $$
Which does not exists as it depends on $\theta$ too so it is not differentiable
Is the way is valid?
Its way too complicated. The easiest is to find the directional derivative $$df_x(v)=\lim\limits_{t\to 0}\frac{f(x+tv)-f(x)}{t}$$
In your case if $v=(a,b)$ you get
$$\frac{f(t(a,b))-f(0)}{t}=\frac{a^3+tb^4}{a^2+b^2}$$
So $$df_0(a,b)=\frac{a^3}{a^2+b^2}$$
And this is NOT linear, thus the function is not differentiable at $0$.