= Solution
{c}
On $(0,1)$, the <product rule> and <chain rule> give $f'(x)=2x\sin(1/x)-\cos(1/x)$. On $(-1,0)$, the <function> is constant and has <derivative> zero. At zero,
$$
\frac{f(h)-f(0)}h=\begin{cases}h\sin(1/h),&h>0,\\0,&h<0.\end{cases}
$$
Its <absolute value> is at most $|h|$, so the <squeeze theorem> gives $f'(0)=0$. Therefore
$$
\boxed{f'(x)=\begin{cases}2x\sin(1/x)-\cos(1/x),&0<x<1,\\0,&-1<x\leq0.\end{cases}}
$$
For $x_n=1/(2\pi n)$, we have $x_n\to0$ but $f'(x_n)=-1$ for every $n$. This differs from $f'(0)=0$, so $f'$ is not <continuous> at zero. Indeed, $y_n=1/((2n+1)\pi)$ gives $f'(y_n)=1$, so the right-hand <limit of a function> does not exist. Thus \b[$f$ is <differentiable> everywhere on $(-1,1)$, while $f'$ is discontinuous at zero.]
Back to article page
