# Thread: A question on infinite derivative.

1. ## A question on infinite derivative.

Definition of infinite derivative: Let $E=\text{dom }f, c\in E$, c is a limit point of E, real-valued function f is said to have a infinite derivative at c if f is continuous at c and the limit $\lim\limits_{x\to c}\frac{f(x)-f(c)}{x - c}$ is $+\infty$ or $-\infty$. We write in this case $f'(c)=+\infty$ or $f'(c)=-\infty$.
For finitely differentiable functions it is well-known that $(f+g)'=f'+g'$. But if both f and g has infinite derivative at c, what about the differentiability of the sum f+g at c? It may be still differentiable, e.g. $f=x^{1/3}$ and $g=-x^{1/3}$, f+g=0 so (f+g)'=0 at c=0 even if $f'(c)+g'(c)=(+\infty)+(-\infty)$ which is undefinable. Now I want to find an example such that $f'(c)=+\infty, g'(c)=-\infty$, but f+g is not differentiable at c, where c is a finite real number. I hope to find a function F satisfying 1)continuous at c, 2)has bounded derivative about c but except c and 3) $F'$ oscillates like $\sin\frac{1}{x}$ so that F is not differentiable at c. If I can find such F, My example can be easily constructed. But I failed to find such F, Can you help me find such F, or construct the example in another way? Thanks!

2. Try two functions that behave more or less the same way as the functions you gave, but don't combine. How about $f(x)=\sqrt[3]{x}$, and $g(x)=-\sqrt[5]{x}$? Here's an example where, I think, there's too much symmetry, and the solution involves destroying symmetry.

3. $\sqrt[3]{x}-\sqrt[5]{x}$ does not meet my requirement: it has derivative $-\infty$ at 0. We should not only destroy the symmetry, but also the destroyer is not differentiable at c. Since derivatives do not have first-kind discontinuities, I think the destroyer should have an oscillating derivative, that is, second-kind discontinuity, near c, as condition 3) in my first post indicated.

4. Wow. It is not intuitive at all that my candidate actually had a derivative at 0. I must confess I didn't actually check it. I just figured you'd have a cusp at the origin, which would not have a definable two-sided derivative. If you plot my function, it does appear to have a cusp at the origin. But it must hang to the left of the origin just enough to get that negative slope.

Anyway. You've got a tough assignment there. I would look into the construction of the everywhere continuous, nowhere differentiable function. I realize that that function, as is, does not solve your problem. But its construction might give you some ideas. You might also try cycloids which have bona fide cusps.