Derivative of differentiable function need not be continuous

Statement

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

Equivalently, a differentiable function on the real numbers need not be a continuously differentiable function.

Proof

Example with an isolated discontinuity

Consider the function:

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

Then, we have:

$g^{\prime }(x)=\{\begin{array}{cc}2x\sin (1/x)-\cos (1/x)& x\ne 0\\ 0,& x=0\\ \end{array}$

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

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

The video below covers this example.

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