 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/1/solution

Apply the chain rule twice: once to the square and once to the argument

x^{3}

of the sine. The inner derivative is

y^{'} = 3 x^{2} cos (x^{3}) - 2 sin (2 x)

. Therefore

\frac{d ( y ^{2} )}{d x} = 2 [sin (x^{3}) + cos (2 x)] [3 x^{2} cos (x^{3}) - 2 sin (2 x)] .

The original PDF has

sin (x^{3})

, rather than the cube of

sin x

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