Real Analysis Derivative Question Help The question is:
Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $g(x)=x^2\sin(1/x^2)$ for $x\neq0$ and $g(0)=0$. Show that $g$ is differentiable $\forall$ $x\in \mathbb{R}$. Also show that the derivative $g'$ is not bounded on the interval $[-1,1]$.
The first part of the question was easy as I just used the definition of the derivative to find $g'(0)=1$. But the second part is bothering me.
$g'(x)=-\frac{2}{x}\cos(1/x^2)+2x\sin(1/x^2)$ for $x\neq0$
To show that this function is unbounded on $[-1,1]$ means that there does not exist a constant $M$ such that $\vert{g'(x)}\vert \leq M$...?
How do you prove the second part of this? Any idea will be helpful to me!
 A: Hint: Consider the points $x_n = \dfrac{1}{\sqrt{2\pi n}}$ for $n \in \mathbb{N}\setminus\{0\}$.
Added Later: To show $g'$ is unbounded on $[-1, 1]$, it is enough to show that for every $M > 0$, there is $x \in [-1, 1]$ such that $|g'(x)| > M$. In particular, if there is a sequence $\{x_n\}$ in $[-1, 1]$ with $|g'(x_n)| \to \infty$, $g'$ is unbounded.
So why the sequence given above? 
First of all, for $x \neq 0$, $g'$ is given by the sum of two terms. The first term is, aside from a constant factor, the product of $\frac{1}{x}$ and $\cos(\frac{1}{x^2})$. We know that $\cos(\frac{1}{x^2})$ is always between $-1$ and $1$; on the other hand, as $x$ gets approaches zero, $\frac{1}{x}$ grows (in absolute value) without bound. The second term is, aside from a constant factor, the product of $x$ and $\sin(\frac{1}{x^2})$, both of which are between $-1$ and $1$. 
What this preliminary analysis tells us is that if $g'$ is unbounded, it must be due to the presence of $\frac{1}{x}$; in particular, the second term is bounded so it doesn't have any effect on the unboundedness of $g'$. With this in mind, we can consider those $x$ for which the second term is zero as this allows us to look at only the first term.
If $\sin(\frac{1}{x^2}) = 0$, $x = \frac{1}{\sqrt{\pi k}}$, $k \in \mathbb{N}\setminus\{0\}$. As $\cos^2\theta + \sin^2\theta = 1$, if $\sin(\frac{1}{x^2}) = 0$, then $\cos(\frac{1}{x^2}) = \pm 1$ and this depends on whether $k$ is even or odd. To simplify matters, we can just consider even $k$ so that $\cos(\frac{1}{x^2}) = 1$; that is, set $k = 2n$ which gives $x = \frac{1}{\sqrt{2\pi n}}$. For such $x$ we have $g'(x) = -\frac{2}{x}$. Now, setting $x_n = \frac{1}{\sqrt{2\pi n}}$, we note that $x_n \to 0^+$, so $g(x_n) \to -\infty$.
