basic math

>>7918250
No, f(x)=|x|

Also, you're not guaranteed there even is a derivative
https://en.wikipedia.org/wiki/Weierstrass_function

>>7918259
suppose the function is differentiable. does it follow in that case?

>>7918263
Nope, standard example is [math]f[/math] defined by [math]f(x) = x^2 \sin(\frac{1}{x}) [/math] for [math] x \neq 0, f(0) = 0 [/math] iirc. That function is differentiable everywhere, but the derivative spazzes out around [math]x = 0[/math].

>>7918263
No. Only if it is infinitely differentiable.

Not OP but what about the antiderivative?

>>7918276
the only point in the derivative that doesn't work is the point that was excluded in the original function. is there a case where the derivative is non-continuous where the original function is bounded?

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>>7918276
neat

>>7918263
Differentiability indeed does imply continuity.

>>7918349
This.

>>7918316
Digital dancing

>>7918349
Not in the case of y=x/x. It's fully differentiable, but there's a hole at x=0, so it's not continuous.

>>7918349
It does, differenciability is defines for points in functions and it must be continuous there.

>>7918316
I actually really like this function because even though it is continuous, it has an infinite amount of cycles within the finite domain [-0.5,0.5], and the period approaches 0 as x approaches 0

>>7918384
x/x isn't differentiable at x = 0, as it's not fucking defined there.

>>7918279
no, that's not a necessary condition.

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>>7918365

Not sure if this tread is real basic shit but it's the first one I consider above basic knowledge that I can also contribute to, nice

>>7918359
>>7918349
>>7918402

Guys, don't mix up words. Yes, differentiability of a function implies continuity of the same function, but he was asking about the derivative. If a function is differentiable, its derivative need not be continuous, as shown in >>7918276 and
>>7918316


>>7918314
I don't really understand your question. If by "original function" you mean [math]f[/math], then the example in >>7918276 is indeed bounded in a neighbourhood of the point at which continuity of the derivative breaks down.

>>7918518
Same here. /sci/ is usually really stupid shit that I don't want to contribute to or really weird shit that I can't contribute to. This one is nice.

>>7918982
Both ff and its derivative are continuous at all points except x=0. I don't see how -2cos(1/x)/x doesn't work outside of x=0.

>>7918316
The cool thing is that you can get continuous derivatives of higher and higher order at the origin by multiplying by higher powers of x.

>>7918279
Nah, twice differentiable is enough (and not even necessary)

>>7918518
You're getting a great chance to see how many people talk out of their asses here as if they're relaying facts.

>>7920193
You're right, this function's derivative is continuous at every point [math]x \neq 0 [/math]. Though we didn't "exclude" anything from the original function, the original function [math]f(x) = x^2 \sin(\frac{1}{x})[/math] possesses a term [math]\frac{1}{x}[/math] which is trouble in [math]x = 0[/math]. Also, the function is not definable in any smooth way other than by setting [math]f(0) = 0[/math]. And, if you're worried that we just "broke" the function somehow by setting the value in [math]x = 0[/math] to something stupid, no, the function is indeed differentiable in this exact point, which you can easily see by writing down the difference quotient at this point.

If you want, you can create functions that have a discontinuous derivative at way more points, even infinitely many. The answers here http://math.stackexchange.com/questions/292275/discontinuous-derivative discuss that. Though the set of points at which you're gonna be able to have a discontinuous derivative is gonna be relatively "small" (even though you can still have infinitely many points in it). They discuss functions like https://en.wikipedia.org/wiki/Volterra%27s_function this one in there.

The basic idea is that you can create discontinuities like the one from our example here on many different points, as long as you're still gonna have a small neighbourhood of these points in which you demand your function's derivative to be continuous, intuitively because you're always gonna need a neighbourhood in which things are going relatively smoothly, otherwise you'll lose differentiability.

Uuuh, yeah. Not sure if I cleared anything up or just confused people.

>>7920321
Seems like quite the cop-out to create a piece-wise function to eliminate discontinuity and then not do the same for the derivative.

>>7920193
just realized that derivative is terribly wrong

>>7918313
>what about the antiderivative?
what about a verb in your question?

>>7920334
Piecewise defined functions aren't really all that special though. I mean, the function, defined in the piecewise way that it is, is still a continuous function, even a differentiable one. But I can't just "do the same" for the derivative, because the derivative is uniquely given by calculating the derivative at every point. I can't mess with it anymore.
And just for clarification, [math]f'(x) = 2x \sin( \frac{1}{x} ) - \cos (\frac{1}{x}) [/math] for [math] x \neq 0 [/math] by product and chain rule, and [math]f'(0) = 0[/math] by just explicitly calculating the difference quotient in this point.