# Thread: Two questions on derivatives

1. ## Two questions on derivatives

1) Assuming f is a real-valued function defined on [a,b] of the real line, f is differentiable on the whole [a,b], is it possible that $\displaystyle \lim\limits_{x \to a} f'(x)$ does not exist (both finite and infinite) while $\displaystyle f'(a)$ exists?
2) Assuming function $\displaystyle f: E\to\mathbb R$, E a subset of real line, let E' be the subset of E on which $\displaystyle f'$ is defined, is it possible that E' contains isolated point?
Thanks.

2. for (1),example: $\displaystyle \text{ If }x\text{(not 0) in }[-1,1],\text{ then let }f(x)=x^2\sin \frac{1}{x},f(0)=0$.
for(2), take f(x) as above mentioned, and $\displaystyle E=\{\frac{1}{z}:z\in Z,z\neq 0\}\cup\{0\}$.

3. Thank you for the first example. Because the formula can not be displayed, I rewrite it as follows: Define $\displaystyle f=\left\{ {\begin{array}{*{20}{c}} {x^2\sin\frac{1}{x}, x\in(0,1]} \\ {0, x=0} \\ \end{array}} \right.$, then f'(0)=0 so f is differentiable on the whole [0,1], but $\displaystyle \lim\limits_{x \to 0} f'(x)$ does not exist (discontinuity of the second kind).
As for your answer to my second question, I think I've got it although some correction may be needed. The domain E should be $\displaystyle \{\frac{1}{z}:z\in\mathbb Z,z\neq 0\}\cup\{0\}$. 0 is the limit point of E, so any limit of function remains the same and f is differentiable at 0 as a result. But sorry for my vague expression in the orignial question, I mean that the isolated point in E' is with regard to E, that is, the isolated point of E, not merely of $\displaystyle \mathbb R$. This is not the case for this example because 0 is a limit point, so not a isolated point, of E.
For clarity I restate the second question as follows:
2) Assuming function $\displaystyle f: E\to\mathbb R$, E a subset of real line, let E' be the subset of E on which $\displaystyle f'$ is defined, is it possible that E' contains element which is an isolated point of E?