calculus - Apparently smooth function, discontinuous derivative

This question is about the existence of discontinuous derivatives, but it doesn't provide much examples except this one and $y = |x|$.

The function

$$f(x) = \left\{ \begin{array}{lr}
\cos(ax) & 0 \leq x \leq c\\
\cos(ac) e^{-b (x - c)} & x > c
\end{array}
\right.$$

has $[0, +\infty)$ as its domain. $a, b, c \in \mathbb (0,+\infty)$ are real constants. Its first derivative is

$$f'(x) = \left\{ \begin{array}{lr}
-a \sin(ax) & 0 \leq x \leq c\\
- b \cos(ac) e^{-b (x - c)} & x > c
\end{array}
\right.$$

$f'(x)$ is continuous only when $a/b = \cot(ac)$ and not in general.

$f(x)$ is a continuous function, without a vertical tangent, infinitely differentiable in its domain, and $a,b$ can be chosen such that the function doesn't have corners in $x = c$; despite this, it has a discontinuous derivative.

1) How can (even graphically) the derivative be not continuous where the function is so?

2) Are there any other similar examples that can be done?

Answer

Disclaimer: This is not exactly an answer to the two questions that you asked, but rather an explanation of what may have caused your confusion.

There two incorrect statements in your question, which may be part of the reason you're confused. First of all, the correct derivative of the function in your example is
$$f'(x) = \left\{\begin{array}{lr} -a\sin(ax) & 0\le x\color{red}{<}c \\ -b\cos(ac)e^{-b(x-c)} & x>c \end{array}\right.$$

Placing "$\le$" instead of the red "$<$" was wrong precisely because in general the derivative of this function at $c$ doesn't exist (except in the special case when $a/b=\cot(ac)$, as you observed).

Because of that, your statement that

$f(x)$ is a continuous function, without a vertical tangent, $\color{red}{\text{infinitely differentiable}}$ in its domain

is false, since in fact this function isn't even differentiable once at $c$ (again, in general).

There's a bit of ambiguity in these discussions of "discontinuous" derivatives. The example that you linked to and $y=|x|$ are two very different things!

• The derivative of $y=|x|$ doesn't exist at $x=0$; and your example is similar to it, since the derivative of your function doesn't exist at $x=c$ either.




• But the linked example, as well as the great answer from @ElliotG, address a much more interesting, but a different question: functions for which the derivative exists everywhere but is discontinuous.