Office_Shredder said: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
[tex] (x^2+y^2) \sin( \frac{1}{\sqrt{x^2+y^2}} ) [/tex]
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)
Whovian said:Please be more specific. What are your calculations, for instance?
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.
The derivative at a point is defined as the slope of the tangent line to a function at that specific point. It represents the instantaneous rate of change of the function at that point.
The limit definition of the derivative at a point is defined as the limit of the change in the function divided by the change in the independent variable, as the change in the independent variable approaches zero. This can be represented mathematically as f'(a) = lim (x->a) (f(x) - f(a)) / (x-a).
A function is continuous at a point if the limit of the function at that point is equal to the value of the function at that point. In other words, the function has no abrupt changes or holes at that specific point.
The function (x^2 + y^2)sin(1/(x^2+y^2)) is continuous at 0 because the limit as (x,y) approaches (0,0) exists and is equal to 0. This means that the function has no abrupt changes or holes at (0,0).
The derivative of (x^2 + y^2)sin(1/(x^2+y^2)) at 0 is equal to 0. This can be found using the limit definition of the derivative, as the limit as (x,y) approaches (0,0) of the function is equal to 0.