 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/1/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

For real

x

arctan x

lies in

(- π /2, π /2)

and has sine

x / 1 + x^{2}

. Thus the principal arcsine in the expression equals the arctangent of

x

\frac{d y}{d x} = \frac{1}{1 + x ^{2}}, x \in R .

The chain rule gives the same result directly, using

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

with a positive square root.

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