I'm currently working on a question in an assignment, and it asks us to use the bounded derivative test, to determine whether $$f(x)=\sin\left(\frac{1}{x}\right)$$ is uniformly continuous on the interval $(0,1)$.
The derivative $f'$ is not bounded; does this imply that $f$ is not uniformly continuous?
Also, in the second part of the question it asks us to do the same with $$f(x)=x \sin\left(\frac{1}{x}\right)$$, I need a worked solution to compare with mine.
No! Actually $$\text{bounded derivative $\Longrightarrow$ uniform continuity } $$ is true with some assumption. . The equivalent true statement is $$\text{non uniform continuity $\Longrightarrow$ non bounded derivative} $$ The statement $$\text{non bounded derivative $\Longrightarrow$ non uniform continuity } $$ is false. For example take $$f:[0,\infty) \ni x \mapsto \sqrt{x} \in \Bbb{R}$$ is uniformly continuous but the derivative is unbounded near zero.
Actually your $f$ is not uniform continuous on $(0,1)$. For,take $$x_n=\frac{1}{\frac{\pi}{2}+2n \pi} \;\; \text{and}\;\;y_n=\frac{1}{\frac{3\pi}{2}+2n \pi}$$ Then $\vert x_n - y_n \vert \rightarrow o$ whereas $\vert f(x_n)-f(y_n) \vert $ is not!
The derivative of the second one is $$f'(x)=\sin (\frac{1}{x})-\frac{1}{x}\cos (\frac{1}{x})$$
Is it bounded? Note that $f$ is uniformly continuous on $(0,1)$