# Differentiability at (0,0).

I always get stuck when I've to show something is differentiable,like in the following question:

$$f(x,y) = \begin{cases} \dfrac{x^3y^3}{x^4+y^4} & \text{if (x,y)\neq(0,0)} \\ 0 & \text{if (x,y)=(0,0)} \end{cases}$$

show that f is differentiable at (0,0) ...

First I thought of showing that partial derivatives exist for the point (0, 0).

## $$\frac{\partial f}{\partial x}(0,0)=\lim_{x\rightarrow 0}\frac{f(x,0)-f(0,0)}{x-0}=\lim_{x\rightarrow 0}\frac{\frac{0}{x^4}}{x}=\lim_{x\rightarrow 0}\frac{{0}}{x^5}= 0$$

later in relation to y:

$$\frac{\partial f}{\partial y}(0,0)=\lim_{y\rightarrow 0}\frac{f(0,y)-f(0,0)}{y -0}=\lim_{y\rightarrow 0}\frac{\frac{0}{y^4}}{y}=\lim_{y\rightarrow 0}\frac{{0}}{y^5}= 0$$

alright both partial derivatives exists and are equal,so now we have to show that :they are continuous near (0,0)..

but that doesn't prove the limit is differentiable, what can I do?

I thought of using the following formula for a corollary.

$$E(h,k)=f(x_0 +h, y_0 +k)-f(x_0, y_0)-\frac{\partial f}{\partial x}(x_0,y_0)h -\frac{\partial f}{\partial y}(x_0,y_0)k$$

however:

$$E(h,k)=\frac{(x_0+h)^3(y_0+k)^3} {(x_0 + h)^4 + (y_0 +k)^4} -\frac{x^3 y^3}{x^4 + y^4}-0h -0k$$

• What definition do you have for differentiability? Dec 12, 2021 at 3:33
• Check your work. $f(x,0)$ is not equal to $x^3/x^4$. Dec 12, 2021 at 3:39
• but in the function $f(x,y)$ is $\frac {x^3 y^3}{x^4+y^4}$
– Skye
Dec 12, 2021 at 3:52
• Is $x\neq 0$ then $f(x,0)=\frac{x^3\cdot 0^3}{x^4+0^4}=0$. What is the definition of differentiability that you're using? Dec 12, 2021 at 4:10
• First I thought of showing that partial derivatives exist for the point (0, 0).
– Skye
Dec 12, 2021 at 12:44

## 1 Answer

As you have correctly computed that $$\frac{\partial f}{\partial x} =\frac{\partial f}{\partial y}=0$$.

A function is said to be differentiable at a point $$(a,b)$$ iff there exist $$\phi(h,k)$$ and $$\psi(h,k)$$ from $$\mathbb{R}^{2}\to\mathbb{R}$$ such that

$$f(a+h,b+k)=h\frac{\partial f}{\partial x}\vert_{(a,b)}+k\frac{\partial f}{\partial y}\vert_{(a,b)}+h\phi(h,k)+k\psi(h,k)$$ such that $$\phi(h,k)\to 0\quad \text{and}\quad\psi(h,k)\to 0$$ as $$(h,k)\to(0,0)$$.

So if our given function is differentiable.

Then $$f(0+h,0+k)=h\frac{\partial f}{\partial x}+k\frac{\partial f}{\partial y}+h\phi(h,k) +k\psi(h,k)$$ where $$\psi$$ and $$\phi$$ both $$\to 0$$ as $$(h,k)\to (0,0)$$ .

Then $$\dfrac{h^3k^3}{h^4+k^4}=h\phi(h,k)+k\psi(h,k)$$.

Take $$\displaystyle\phi(h,k)=\frac{h^{2}k^{3}}{2(h^{4}+k^{4})}$$ and $$\displaystyle\psi(h,k)=\frac{k^{2}h^{3}}{2(h^{4}+k^{4})}$$.

Then both $$\phi$$ and $$\psi$$ tend to $$0$$ as $$(h,k)\to(0,0)$$.

Hence you have differentiability.

In a Equivalent formulation of Differentiability.

A function is differentiable at a point $$(a,b)$$ iff there exist $$\psi(h,k)$$

$$f(a+h,b+k)=h\frac{\partial f}{\partial x}\vert_{(a,b)}+k\frac{\partial f}{\partial y}\vert_{(a,b)}+\sqrt{h^{2}+k^{2}}\cdot \psi(h,k)$$ such that $$\psi(h,k)\to 0$$ as $$(h,k)\to(0,0)$$.

In this case you just take $$\displaystyle\psi=\frac{h^3k^3}{(h^4+k^4)\sqrt{(h^{2}+k^{2})}}$$ . Then this would also imply differentiablity.

It's just a matter of what convention you follow. Many people like this norm formulation. I myself prefer the previous one. They are both equivalent.

Lastly, I would say that you should try and use polar coordinates to show that $$\psi$$ and $$\phi$$ tend to $$0$$. It is very easy. $$\cos^{4}(\theta)+\sin^{4}(\theta)\geq \frac{1}{2}$$ always.

• Could you tell me why in the corollary you added it all up and what does the: $\frac{\partial f}{\partial x}h |_(a,b)$ and what does it mean ϕ and ψ?
– Skye
Dec 12, 2021 at 14:37
• @Skye $\frac{\partial f}{\partial x}|_{(a,b)}$ just means evaluated at $(a,b)$. Nothing else. And $\phi$ and $\psi$ are two functions from $\mathbb{R}^{2}\to\mathbb{R}$ such that both tend to $0$ as $(h,k)\to(0,0)$ Dec 12, 2021 at 14:39
• Got it, can you show me where I can study a formula you presented at the beginning?
– Skye
Dec 12, 2021 at 14:45
• It's just a formulation of differentiation. Many people also wirte it as $\frac{f(a+h,b+k)-hf_{x}-kf_{y}}{\sqrt{h^{2}+k^{2}}}\to 0$ as $(h,k)\to (0,0)$. Here $f_{x}$ means partial derivative. I wouls suggest you the book by Satish Shirali and Harikrishna Lal Vasudeva on multivariable calculus for this reference. It is frankly imo the best book on the topic and it helped me a lot at the undergrad level. Dec 12, 2021 at 14:48
• Also @Skye you can mark as answer if it has answered your question Dec 12, 2021 at 14:49