Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by
$$
f(x):=\left\{\begin{array}{ll}
x^{2} \sin \left(1 / x^{2}\right) & \text { if } x \neq 0 \\
0 & \text { if } x=0
\end{array}\right.
$$
Show that $f$ is differentiable on $\mathbb{R}$, but for any $\delta>0, f^{\prime}$ is not bounded on $[-\delta, \delta] .$ Thus $f^{\prime}$ has an antiderivative on the interval $[-1,1]$, but it is not Riemann integrable on $[-1,1]$.