Solution
= Solution
Apply the <product rule> to $x\sin(1/x)$ and the <chain rule> to the <sine>. Since $d(1/x)/dx=-1/x^2$, the <derivative> is
$$
\boxed{\frac{dy}{dx}=\sin\frac1x-\frac1x\cos\frac1x,\qquad x\ne0.}
$$
= Solution
Apply the <product rule> to $x\sin(1/x)$ and the <chain rule> to the <sine>. Since $d(1/x)/dx=-1/x^2$, the <derivative> is
$$
\boxed{\frac{dy}{dx}=\sin\frac1x-\frac1x\cos\frac1x,\qquad x\ne0.}
$$