Study this function

1. Dec 8, 2018

Kenneth1997

1. The problem statement, all variables and given/known data
study the continuity, directional derivatives, and differentiability of the function f(x,y)=arctan(abs(y)*(y+x^2-1)).

3. The attempt at a solution
the function is obviously continuous in R2 since made of continuous functions.

has directional derivatives everywhere since made of functions that has directional derivatives everywhere.

differentiable everywhere exept for (x,0) and here is my biggest doubt: how do i demonstrate that it isnt differentiable there? if i think about it on a logical level, i know these are point where the function isnt smooth, but how do i demonstrate it?

in (+-1,0) its differentiable because for h=x+-1,k=1:
lim(h,k)->(0,0) arctan(abs(k)*(k+h^2+-2h))/(sqrt(h^2+k^2)) goes obviously to 0.

Last edited: Dec 8, 2018
2. Dec 8, 2018

Ray Vickson

As written your function has unbalanced parentheses, so could represent either
$$f = \left(\arctan |y| \right)(y+x^2-1) \hspace{4ex}(1)$$
or
$$f = \arctan(|y|(y+x^2-1)) \hspace{4ex}(2)$$
I suspect you mean (2), but (1) is a perfectly credible way of filling in the missing information.

3. Dec 8, 2018

Kenneth1997

edited,yep i meant the second

4. Dec 8, 2018

Ray Vickson

The "inside" function $g(x,y) + |y| (y+x^2-1)$ can be written as $g(x,y) = y |y| + (x^2-1) |y|.$ The first function--- $y |y|$ --- is differentiable in $y$ (even at $y = 0$) but the second one --- $(x^2-1) |y|$ --- is not differentiable in $y$ at $y = 0, x \neq \pm1.$ So, your function is differentiable at $(1,0)$ and $(-1,0)$ but not at any other $(x,0).$

5. Dec 8, 2018

Kenneth1997

but how do i demonstrate it?

6. Dec 8, 2018

Ray Vickson

Helpers are not allowed to show you all the details. I have said as much as I can (maybe even more than I should have) under PF rules.

7. Dec 8, 2018

Kenneth1997

a better understanding should be the goal not solving homework to high schoolers imo. i already demonstrated what you did so i dont think you are showing more then you should have.