 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019/5/solution

g (x) = 8 x + sin^{2} (2 x)

. The chain rule gives

f^{'} = - g^{'} / g^{2}

, while

g^{'} (x) = \frac{4}{8 x} + 4 sin (2 x) cos (2 x) = \frac{4}{8 x} + 2 sin (4 x) .

Thus

f^{'} (x) = - \frac{4/ 8 x + 2 sin ( 4 x )}{[ 8 x + sin ^{2} ( 2 x ) ] ^{2}}, x > 0.

The real domain excludes zero because the original denominator vanishes there; for

x > 0

it is strictly positive.

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