differentiability on closed interval

Question

Let $f(x)=x^2$ be defined only on [0,1], is $f$ differentiable at $x=0$ and $x=1$? On the one hand it is obviously differentiable and $f'(x)=2x$, on the other hand one sided limits of difference quotients are undifined, for example: $\lim_{h \to 0^{-}}\frac{f(x+h)-f(x)}{h}$, because $f$ does not exist outside $[0,1]$

Answer 1

This is up to one's definition of differentiability. For instance, if I recall correctly, the calculus textbook my university uses only requires a one-sided derivative at the endpoints, i.e. $f : [a,b] \to \mathbb{R}$ $(a<b)$ is defined to be differentiable on $[a,b]$ if these three are satisfied:

• $f'(x) := \displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists for all $x \in (a,b)$
• $f'(a) := \displaystyle \lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}$ exists
• $f'(b) := \displaystyle \lim_{h \to 0^-} \frac{f(b+h)-f(b)}{h}$ exists

I also feel like I've seen textbooks not allow differentiability on the endpoints as well, so it is best to consult your source material in further detail or ask your instructor (so everyone's on the same page).