Answer
No, Yes; see explanations.
Work Step by Step
Step 1. For the derivative of the function
$h(x)=\begin{cases} x^2sin\frac{1}{x},\ x\ne0\\0,\ x=0 \end{cases}$
to be continuous at $x=0$, we need to show that $lim_{x\to0}h'(x)=$ exists. (We have shown the function is differentiable at $x=0$ in Exercise 22-d).
Step 2. $lim_{x\to0}h'(x)=lim_{x\to0}(2x\ sin\frac{1}{x}+x^2\ cos\frac{1}{x}(-\frac{1}{x^2}))=lim_{x\to0}2x\ sin\frac{1}{x}-lim_{x\to0}\ cos\frac{1}{x}$.
Step 3. As $cos\frac{1}{x}$ oscillates when $x\to0$, the second limit from above does not exist. Thus, the derivative of $h(x)$ is not continuous at $x=0$.
Step 4. For $k(x)=xh(x)$, $lim_{x\to0}k'(x)=lim_{x\to0}(3x^2\ sin\frac{1}{x}+x^3\ cos\frac{1}{x}(-\frac{1}{x^2}))=lim_{x\to0}3x^2\ sin\frac{1}{x}-lim_{x\to0}x\ cos\frac{1}{x}=0-0=0$. (because $|cos\frac{1}{x}|\leq1$). Thus, the derivative of $k(x)$ is continuous at $x=0$.
