Study Function: Continuity, Derivatives & Differentiability

[Kenneth1997]

Homework Statement

study the continuity, directional derivatives, and differentiability of the function f(x,y)=arctan(abs(y)*(y+x^2-1)).

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 isn't differentiable there? if i think about it on a logical level, i know these are point where the function isn't 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.

 

[Ray Vickson]

Homework Statement

study the continuity, directional derivatives, and differentiability of the function f(x,y)=arctan(abs(y)*(y+x^2-1).

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 isn't differentiable there? if i think about it on a logical level, i know these are point where the function isn't 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.

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.

 

[Kenneth1997]

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.

edited,yep i meant the second

 

[Ray Vickson]

Homework Statement

study the continuity, directional derivatives, and differentiability of the function f(x,y)=arctan(abs(y)*(y+x^2-1)).

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 isn't differentiable there? if i think about it on a logical level, i know these are point where the function isn't 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.

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 $(x,0).$

 

[Kenneth1997]

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 $(x,0).$

but how do i demonstrate it?

 

[Ray Vickson]

but how do i demonstrate it?

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.

 

[Kenneth1997]

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.

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