# Thread: Continuity, differentiability and partial derivatives

1. ## Continuity, differentiability and partial derivatives

Let $n=0,1,2,\ldots$ and $f:\mathbb R^2\to\mathbb R$ defined by

$f(x,y)=\left\{\begin{array}{cl}(x+y)^n\sin\dfrac1{ \sqrt{x^2+y^2}}&(x,y)\ne(0,0).\\[0.5cm]0&(x,y)=(0,0).\end{array}\right.$

How should we choose $n$ such that:

a) $f$ is continuous.

b) $f$ is differentiable.

c) $f$ has continuous partial derivatives on every point on $\mathbb R^2.$

Spoiler:

a) is easy, I found that for $n\ge1$ we have the continuity.

As for b), I'm not sure, do I just need that $\displaystyle\lim_{h\to0}\frac{f(x+h,y)-f(x,y)}h$ and $\displaystyle\lim_{h\to0}\frac{f(x,y+h)-f(x,y)}h$ exist? But, do they need to be equal? I think the latter is not necessary.

And c), I'd have to study $\displaystyle\lim_{(x,y)\to(x_0,y_0)}f_x,$ in the same fashion for $f_y,$ is this correct?

2. For a) you're right although maybe it would be more appropiate to write $n>0$ in case you decided to let $n\in \mathbb{R}$. For b) you need to verify whether there exists a linear transformation $L:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $\lim_{h\rightarrow 0} \frac{|f(0+h)-f(0)-L(h)|}{\| h\| }=0$. For c) you're right.

Hint:
Spoiler:
a) n>0, b) n>1, c) n>2

3. Okay, I got a) and c), but as for b), how do I find the linear transformation?

Wow, that's new to me, but, is my approach for b) correct or something?

4. Originally Posted by Killer
Wow, that's new to me, but, is my approach for b) correct or something?
Kind of, it will give you that the function can't be differentiable if $n\leq 1$, but to prove differentiablity in case n>1 you'll need the definition I gave you. Before looking at the hint, notice that in case n>1 the function approaches zero very fast near the origin (try to picture it in one dimension) so the derivative must be...

Hint:
Spoiler:
take $L\equiv 0$