 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-1/3e/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

No. Take

f (x) = ∣ x ∣

a = 0

, and

g (y) = y^{2}

. Then

f

is not differentiable at zero, but

g \circ f = x^{2}

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