 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/11/a/i/solution

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

y^{'} = \frac{( 4 x ^{3} + 3 x ^{2} ) ( 3 x - 2 ) - 12 x ^{3} ( x + 1 )}{( 3 x - 2 ) ^{5}} .

The numerator simplifies to

- 11 x^{3} - 6 x^{2}

y^{'} = - \frac{x ^{2} ( 11 x + 6 )}{( 3 x - 2 ) ^{5}}, x \neq = \frac{2}{3} .

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