If I am asked to prove that the function $x^3 \sin(\frac{1}{x})$ is differentiable for all $x \ne 0$, is it sufficient to say something like the following:
We see that $x^3$, $\sin(\frac{1}{x})$ and $\frac{1}{x}$ are differentiable for all $x \ne 0$. It follows, therefore, that their product is also differentiable for all $x \ne 0$.