Derivative of differentiable function need not be continuous

This article gives the statement and possibly, proof, of a non-implication relation between two function properties. That is, it states that every function satisfying the first function property (i.e., differentiable function) need not satisfy the second function property (i.e., continuously differentiable function)
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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.

See also

• Difference quotient: We can relate the continuity of the derivative to the joint continuity of the difference quotient.

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.