Example of function that is differentiable, but the second derivative is not defined Is there an example of function that is differentiable at $a$, but the second derivative is not defined at $a$? I bet that this is not possible, because if the function is differentiable then it is smooth and the slope is not vertical, so the only way the second derivative to be not defined is if the slope of the first derivative is vertical. That means that the rate of change is ambiguous at $a$, since there are at least two points almost equal for the same $x$, so I bet that such function doesn't stand a chance not even in the real world, but also in the wonderword of math. Prove me wrong!
 A: $f(x) = x^\frac{5}{3}$ is such an example.  The domain is all real numbers.
Then, $f'(x)$ is essentially $f(x) = x^\frac{2}{3}$, which also has domain all real numbers.
Then, $f"(x)$ is essentially $f(x) = x^\frac{-1}{3}$, which has domain all real numbers except 0.  
So for this given function $f(x)= x^\frac{5}{3}$, $f'(0)$ is defined but $f"(0)$ is NOT defined.
A: The Weierstrass function is a famous example of a function which is everywhere continuous, but nowhere differentiable.  

Let us write $f(x)$ for this function.  Then the function given by
$$
F(x)=\int_0^xf(t)dt
$$
is differentiable everywhere but twice differentiable nowhere.  
See this answer for graphs of both functions.
A: A smooth function is a function that is infinitely often differentiable at every point of its domain. However, not every differentiable function satisfies this condition.
For example, the function $f(x) = \begin{cases}0 & x = 0 \\ x^2 cos\left(\frac{1}{x}\right) & \text{else} \end{cases}$ is everywhere differentiable, but the derivative isn't even continuous in 0.
