Study Function: Continuity, Derivatives & Differentiability

In summary, the function f(x,y)=arctan(abs(y)*(y+x^2-1)) is continuous in R2 and has directional derivatives everywhere. However, it is only differentiable at (1,0) and (-1,0), but not at any other point (x,0). This is because the "inside" function has a differentiable part in y, but also a non-differentiable part in y at y=0 for x≠±1. It is important to focus on understanding the concept rather than just finding the solution to the homework problem.
  • #1
Kenneth1997
21
2

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.
 
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  • #2
Kenneth1997 said:

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.
 
  • #3
Ray Vickson said:
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
 
  • #4
Kenneth1997 said:

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).##
 
  • #5
Ray Vickson said:
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?
 
  • #6
Kenneth1997 said:
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.
 
  • #7
Ray Vickson said:
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.
 

FAQ: Study Function: Continuity, Derivatives & Differentiability

What is continuity?

Continuity is a mathematical concept that describes the smoothness and connectedness of a function. A function is considered continuous if its graph has no breaks or jumps, and if its output values change continuously as the input values change. Simply put, a function is continuous if you can draw its graph without lifting your pencil.

What is a derivative?

A derivative is a mathematical tool used to describe the rate of change of a function at a specific point. It is defined as the slope of the tangent line to the graph of a function at that point. In other words, it tells us how much the output of a function changes when the input changes by a very small amount.

How do you find the derivative of a function?

The derivative of a function can be found using the rules of differentiation, which involve taking the limit of the slope of a secant line as the two points on the line get closer and closer together. There are also various differentiation rules for different types of functions, such as power functions, exponential functions, and trigonometric functions.

What is differentiability?

Differentiability is a property of a function that means it has a derivative at every point in its domain. A function is differentiable if its graph is smooth and has no sharp turns or corners. This concept is closely related to continuity, as a function must be continuous to be differentiable.

Why are continuity, derivatives, and differentiability important?

These mathematical concepts are essential for understanding and analyzing functions in various scientific disciplines, such as physics, engineering, economics, and more. They allow us to model and predict real-world phenomena, optimize systems, and make informed decisions based on data. Additionally, they form the foundation for more advanced mathematical concepts and techniques.

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