# Continuity, partial derivatives and differentiability of a multivariable function

I am studying the following function: $$f(x,y)=\begin{cases} \frac{x^4y-xy^4}{x^4+y^4}, & (x,y)\neq (0,0)\\ 0, & (x,y)=(0,0) \end{cases}$$ to see if it is continuous, if the partial derivatives exist, compute them, see if they are continuous and if $$f$$ is differentiable.

What I have done:

The domain of $$f$$ is $$\mathbb{R^2}$$.

$$f$$ is clearly continuous in $$\mathbb{R^2}\setminus\{(0,0)\}$$.

The partial $$x$$ derivative $$\frac{\partial f}{\partial x}(x,y)=\frac{3x^4y^4+4x^3y^5-y^8}{(x^4+y^4)^2}$$ is continuous in $$\mathbb{R^2}\setminus \{(0,0)\}, \frac{\partial f}{\partial x} (0,0)=\lim_{t\to 0}\frac{f(0+t,0)-f(0,0)}{t}=0$$ but $$\frac{\partial f}{\partial x}(x,0)=0$$ and $$\frac{\partial f}{\partial x}(x,x)=\frac{3}{2}$$ so it is not continuous in $$(0,0)$$.

The partial $$y$$ derivative $$\frac{\partial f}{\partial y}(x,y)=\frac{x^8-3x^4y^4-4x^5y^3}{(x^4+y^4)^2}$$ is continuous in $$\mathbb{R^2}\setminus \{(0,0)\}, \frac{\partial f}{\partial y} (0,0)=\lim_{t\to 0}\frac{f(0+t,0)-f(0,0)}{t}=0$$ but $$\frac{\partial f}{\partial y}(x,0)=1$$ and $$\frac{\partial f}{\partial x}(x,x)=-\frac{3}{2}$$ so it is not continuous in $$(0,0)$$.

Now, to see if $$f$$ is continuous in $$(0,0)$$ I have tried to compute the $$\lim_{(x,y)\to (0,0)} f(x,y)$$ in various ways (like using polar coordinates) but to no avail, so I would like an hint about how to do this, thanks.

$$|\frac{x^4y-xy^4}{x^4+y^4}|\overset{x=\rho\cos\theta\\ y=\rho\sin\theta}{=}\frac{\rho^5 |\cos^4(x)\sin(x)-\cos(x)\sin^4(x)|}{\rho^4 (\cos^4(x)+\sin^4(x))}\leq \rho\ \frac{2}{\min(\cos^4(x)+\sin^4(x))}\overset{\rho\to 0}{\to} 0$$ so, since $$\lim_{(x,y)\to (0,0)} |f(x,y)|=0$$ we have that $$\lim_{(x,y)\to (0,0)}f(x,y)=0$$ and $$f$$ is thus continuous at the origin too.

$$\lim_{(h,k)\to (0,0)}\frac{f(0+h,0+k)-f(0,0)-\vec{\nabla}f(0,0)\cdot (h,k)}{\sqrt{h^2+k^2}}=\lim_{(h,k)\to (0,0)}\frac{h^4k-hk^4}{h^4+k^4}\cdot\sqrt{h^2+k^2}=\lim_{\rho\to 0}\rho^2 \frac{\cos^4(\theta)\sin(\theta )-\cos(\theta)\sin^4(\theta)}{\cos^4(\theta)+\sin^4(\theta)}=0$$ so $$f$$ is differentiable at the origin.

• Ah, you changed the definition of $f$ ;-) May 27 at 23:14
• @Paul Frost Yes, the denominator shouldn't have been squared so I corrected $f$ and solved the problem (correctly, I hope). May 27 at 23:27
• Yes, it is correct. May 27 at 23:27

Hint: What is $$\lim_{x\to 0} f(x,-x)$$?