= Solution
{c}
\b[No.] On $(-1,1)$ take $c=0$ and
$$
f(x)=\begin{cases}x^2\sin(1/x),&x\ne0,\\0,&x=0.\end{cases}
$$
At zero, its <difference quotient> has <absolute value> $|x\sin(1/x)|\leq|x|\to0$, so $f'(0)=0$. Away from zero, the <product rule> and <chain rule> apply. Define
$$
A_n=2\pi n,\qquad B_n=\begin{cases}\pi/2,&n\text{ even},\\3\pi/2,&n\text{ odd},\end{cases}\qquad x_n=\frac1{A_n},\quad y_n=\frac1{A_n+B_n}\qquad(n\geq1).
$$
These are distinct positive inputs tending to zero. Since $\sin A_n=0$ and $\sin(A_n+B_n)=\sin B_n$,
$$
\frac{f(y_n)-f(x_n)}{y_n-x_n}=-\frac{\sin B_n}{B_n}\frac{A_n}{A_n+B_n}.
$$
The even-indexed <subsequence> tends to $-2/\pi$, whereas the odd-indexed <subsequence> tends to $2/(3\pi)$. Thus the <limit of a sequence> does not exist. The two points lie on the same side of $c$, so the nonnegative-weight estimate in part (b) is unavailable. This is a concrete instance of <differentiability does not imply continuity of the derivative>.
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