Question #a72ca

Feb 18, 2016

It is false.

Explanation:

For instance $| x |$ is continuous, but does have derivate in $x = 0$.

Additionally to the continuity, if have to guarantee that $f ' \left({c}^{-}\right) = f ' \left({c}^{+}\right)$

Feb 19, 2016

Another example is $f \left(x\right) = \sqrt[3]{x}$
$f \left(x\right) = \sqrt[3]{x}$ is continuous at $0$, but $f ' \left(x\right) = \frac{1}{3 \sqrt[3]{{x}^{2}}}$ is not defined at $0$.