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Let f(x)=∫ from 2x to sin x of cos(t³) dt. Then the value of f''(x) is

1. cos(sin³ x) cos x − 2 cos(8x³)
2. sin(sin³ x) sin x − 2 sin(8x³)
3. cos(cos³ x) cos x − 2 cos(x³)
4. cos(sin³ x) − cos(8x³)

Correct answer: cos(sin³ x) cos x − 2 cos(8x³)

Solution

With f(x)=int_{2x}^{sin x} cos(t^3) dt, the Leibniz rule gives f'(x)=cos(sin^3 x)*cos x - cos((2x)^3)*2 = cos(sin^3 x)cos x - 2cos(8x^3), which is option 1.

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