Derivative of differentiable function need not be continuous

Statement

It is possible to have a function $f$ defined for real numbers such that math>f</math> is a differentiable function everywhere on its domain but the derivative $f^{\prime }$ is not a continuous function.

Proof

Example with an isolated discontinuity

Consider the function:

$f(x):=\{\begin{array}{cc}{x}^2\sin (1/x),& x\ne 0\\ 0,& x=0\\ \end{array}$

Then, we have:

Failed to parse (unknown function "\begin{array}"): {\displaystyle f'(x) = \left\lbrace\begin{array}{rl} 2x \sin(1/x) - \cos(1/x) & x \ne 0 \\ 0, & x = 0 \\\end{arrray}\right.}

In particular, we note that $f^{\prime }(0)=0$ but $\underset{x\to 0}{lim}{f}^{\prime }(x)$ does not exist. Thus, $f^{\prime }$ is not a continuous function at 0.

For details, see square times sine of reciprocal function#First derivative.