 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/3/solution

Apply the product rule to

x sin (1/ x)

and the chain rule to the sine. Since

d (1/ x) / d x = - 1/ x^{2}

\frac{d y}{d x} = sin \frac{1}{x} - \frac{1}{x} cos \frac{1}{x}, x \neq = 0.

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