Degenerate maths functions

>>8484804
>continous and differentiable

This is redundant.

Regarding your question, if a function is differentiable then set of points where the derivative of said function is continuous is non-empty.

This is obvious if you approach it by contradiction.

Suppose you have a function that is differentiable on the interval (a.b) such that its derivative is not continuous in any point of (a.b)

Any discontinuity on this derivative would imply a sudden change in the original function and through some more rigorous analysis you would get that your original function is actually not differentiable anywhere in (a.b), which strongly contradicts the hypothesis.

>>8484827

B...B...But there are differentiable functions which verifies the following proporties :
- the set of points where the derivative is positive is dense
- the set of points where the derivative is negative is dense

>>8484804
http://math.stackexchange.com/questions/1444138/differentiable-functions-such-that-the-derivative-is-nowhere-continuous

sin(x)

because whenever I see it, i read it as "sex"

very degenerate

>>8484931

It is not very clear. What means "continuous on a dense subset" ?

For instance [math] f : x \mapsto 1 [/math] if [math] x \in \mathbf{Q} [/math] and 0 otherwise is continuous on [math] \mathbf{Q} [/math] ? Even if it is continuous nowhere ?

>>8486350
Continuity is defined pointwise: so "continuous on a dense subset" means that the set of points at which the function is continuous is dense.

The function in your example is not continuous at any point of Q. Because for any rational number q, we can find a sequence x_n of irrational numbers that converges to q.

Then [math]\lim_{n\to\infty}f(x_n)=lim_{n\to\infty}0=0 != f(\lim_{n\to\infty}x_n)=f(q)=1[/math]