Limit of derivative

1. Apr 30, 2012

brydustin

obviously x^2sin(1/x) isn't continuous in its derivative at zero. But it seems if you take
(x^2 + y^2) sin(1/(x^2+y^2)) and switch to polar coordinates then the derivative is continuous at zero.... but that can't be right... can it?

2. Apr 30, 2012

Office_Shredder

Staff Emeritus
The second function doesn't restrict to the first function on the x-axis, so I'm not sure why you would be surprised their differentiability properties would be different. On the other hand the function

$$(x^2+y^2) \sin( \frac{1}{\sqrt{x^2+y^2}} )$$
is the radial extension of your original function to the x-y plane, and when you switch to polar coordinates you get that it is not differentiable at zero (since it's the exact same function once you do that)

3. Apr 30, 2012

Whovian

Also, the derivative in polar coordinates aren't that simple. If we have r=f(θ), then f'(θ) isn't necessarily the slope of the line tangent to f at θ.

4. Apr 30, 2012

brydustin

g'(r,alpha) = (for all alpha) = z2r*sin(1/r*sqrt(z)) - r^2/sqrt(r*z) cos(1/r*sqrt(z)) is what I get. where z = cos^2(alpha) + sin^2(alpha). And the limit should exist as r goes to zero by the squeeze theorem., .... right?

5. Apr 30, 2012

brydustin

OH, no.... I made a mistake. I made a bad substitution as I forgot that r = sqrt(x^2+y^2) and then its really trivial. Funny, that I had this idea but did the transformation wrong, still.

Last edited: Apr 30, 2012