2

I came across this theorem in calculus: If fx and fy exist near (a,b) and are continous at (a,b) then f(x, y) is differentiable at (a,b)

What confuses me is that when I look at solutions to questions that require you to use the above theorem, the solutions only find fx and fy and determine if they are continous at (a,b) but they don't show that those partial derivatives exist near (a,b) as well. I want to know if finding fx and fy is and showing they are continous at (a,b) is enough and why?

Example question: Given f(x, y) = (x + 3y)^(1/2). Is the function differentiable at (1, 2).

rert588
  • 402
  • 2
    The partial derivatives cannot be continuous without existing. – Martin R Feb 22 '18 at 12:08
  • Correct me if I'm wrong but can the partial derivatives exist at (a,b) but not near(a,b) – rert588 Feb 22 '18 at 12:09
  • That is correct. If a function is only defined at the (isolated) point (a, b) then it is trivially continuous. – Perhaps you can give a concrete example of a solution that you found, where you think that only continuity at (a,b) but not existence in a neighborhood of (a,b) has been proven. – Martin R Feb 22 '18 at 12:14

2 Answers2

0

Since $f_x(x,y)=(x+3y)^{-1/2}$ and $f_y(x,y)=3(x+3y)^{-1/2}$, both of which are continuous at $(1,2)$, $f$ is differentiable there.

José Carlos Santos
  • 427,504
0

As for the example question:

Since we have: $$f(x,y)=\sqrt{x+3y}$$ we take: $$\begin{align*} f_x(x,y)&=\frac{1}{2\sqrt{x+3y}}\\ f_y(x,y)&=\frac{3}{2\sqrt{x+3y}} \end{align*}$$ which are both continuous - and, of course, well-defined - for every $(x,y)\in\mathbb{R}^2$ such that $x+3y>0$. So, since $(1,2)$ is such a point then abovementioned the theorem gives us that $f$ is differentiable at $(1,2)$.

As for your comment about the existence of partial derivatives in an area around $(a,b)$:

Consider the following function $f:\mathbb{R}^2\to\mathbb{R}$: $$f(x,y)=\left\{\begin{array}{ll} y\left|x^2\sin\frac{1}{x}\right| & x\neq0\\ 0 & x=0 \end{array}\right.$$ Also, let $$A:=\left\{(x,y)\in\mathbb{R}^2\left| x\neq\frac{1}{2k\pi},\ k\in\mathbb{Z}\right.\right\}$$ It is clear that for $(x,y)\in A$, $f_x(x,y)$ does not exist, however, it does exist in $(0,y)$, for every $y\in\mathbb{R}$. Indeed, $$\begin{align*} \lim_{x\to0}\frac{f(x,y)-f(0,y)}{x-0}&=\lim_{x\to0}\frac{y\left|x^2\sin\frac{1}{x}\right|}{x}=\\ &=y\lim_{x\to0}\frac{\left|x^2\sin\frac{1}{x}\right|}{x}=\\ &=y\cdot0=\\ &=0 \end{align*}$$ since $$\left|\frac{\left|x^2\sin\frac{1}{x}\right|}{x}\right|\leq\left|x\right|\Leftrightarrow-|x|\leq\frac{\left|x^2\sin\frac{1}{x}\right|}{x}\leq|x|$$ So, $f_x(0,y)=0$, for every $y\in\mathbb{R}$ but $f_x$ does not exist in any are around $(0,y)$ since it does not exist in $A$ and all points $(0,y)$ are accumulation points of $A$.

Edit:

You can now take $$g(x,y)=f(x,y)+f(y,x)$$ and have the same result for both $g_x$ and $g_y$.

Vassilis Markos
  • 4,484