Everywhere Differentiable? Not Quite

Calculus Level 2

The graph $\left(x^2-y^2\right)^2=x^2+y^2$ is everywhere differentiable except for some points. What is the sum of the coordinates of these points?

The answer is 0.

1 solution

Daniel Liu
May 28, 2014

Note that the graph is symmetric because all terms are $x^2$ and $y^2$ . Thus, for any non-differentiable point, there is an opposite point such that the sum of their coordinates is $0$ . thus, the answer must be $0$ .

A very troll problem I decided to make. The actual points that are not differentiable are $(-1,0),(0,0),(1,0)$

Ii was hard for me to be sure enough to enter that answer. Trolls keep going.....

Nishant Sharma - 7 years ago

Sorry, this was just a for-fun problem, not a really serious one.

Daniel Liu - 7 years ago

@Daniel Liu , I have a doubt. This is a one-many relation. How would you define the slope at a point where a value of x corresponds to two values of y?

Siddharth Brahmbhatt - 7 years ago

Slightly more difficult version: Without finding the actual x-nondifferentiable points of this curve on the complex plane, prove that they must all have radius 1 or $1/\sqrt{2}$ .

Michael Lee - 6 years, 10 months ago

Everywhere Differentiable? Not Quite

The answer is 0.

1 solution

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