Discuss the differentiability of $y=\sin(\pi(x-[x]))$ in $(-\frac{\pi}{2},\frac{\pi}{2})$, where $[x]$ is the largest integer in x which is $\leq x$.
$ y = \left\{ \begin{array}{ll} \sin(\pi(x+2)), & \quad -\frac{\pi}{2}
Discuss the differentiability of $y=\sin(\pi(x-[x]))$ in $(-\frac{\pi}{2},\frac{\pi}{2})$, where $[x]$ is the largest integer in x which is $\leq x$.
$ y = \left\{ \begin{array}{ll} \sin(\pi(x+2)), & \quad -\frac{\pi}{2}
Hint: Because $\sin (x+2k\pi)=\sin x$ and $\sin(x+2k\pi+\pi)=-\sin x$ ($k\in \mathbb{Z}$), your function is in fact: $f(x)=\begin{cases}\sin(\pi x)&\mbox{if, }[x]\text{ is even}\\ -\sin (\pi x)&\mbox{if, }[x]\text{ is odd} \end{cases}$ or since we are working in $(-\frac{\pi}{2},\frac{\pi}{2})$, $f(x)=\begin{cases}\sin(\pi x)&\mbox{if, }0\le x<\frac{\pi}{2}\\ -\sin (\pi x)&\mbox{if, }-\frac{\pi}{2}< x<0 \end{cases}$ I think you can finish the rest if you remember that $\lim_{h\to 0}\frac{\sin ah}{h}=a$
EDIT: Here is the rest:
Obviously $f$ is differentiable in $(-\frac{\pi}{2},0)\cup(0,\frac{\pi}{2})$. We need to check differentiability at $0$ separately with the definition:
$\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^+}\frac{\sin(\pi h)}{h}=\pi$ while $\lim_{h\to 0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^-}\frac{-\sin(\pi h)}{h}=-\pi$ and therefore $f$ is not differentiable at $0$ even though it is continuous there.