I've seen an answer or two on this, but they don't fully make sense with me. One example is that $sin\frac{1}{x}$ is not differentiable at $x=0$, but then isn't the derivative: $$-\frac{cos\frac{1}{x}}{x^2}$$
This is not continuous at $x=0$ as well. Can someone explain this, or give an example of when the differentiability of a function is not the same as the continuity of the derivative?
Best Answer
The answer is no.
Consider the function given by \begin{align*} f(x) = \begin{cases} x^{2}\sin\left(\dfrac{1}{x}\right) & \text{if} \ \ x \neq 0,\\\\ 0, & \text{if} \ \ x = 0 \end{cases} \end{align*}
Such function is differentiable at $x = 0$, but its derivative is not continuous at this point.
Indeed, one has that \begin{align*} f'(0) & = \lim_{x\to 0}\frac{x^{2}\sin\left(\frac{1}{x}\right)}{x} = \lim_{x\to 0}x\sin\left(\frac{1}{x}\right) = 0 \end{align*}
On the other hand, we do also have that \begin{align*} f'(x) = \begin{cases} 2x\sin\left(\dfrac{1}{x}\right) - \cos\left(\dfrac{1}{x}\right), & \text{if} \ \ x \neq 0,\\\\ 0, & \text{if} \ \ x = 0. \end{cases} \end{align*}
which confirms our claim.
Hopefully this helps !