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\(\frac{d}{dx}[\frac{1}{2}sin^2x]\) =d/dx[(1/2)sin2x] =(A) sin2x(B) \(\frac{1}{2}\)sin2x(C) \(\frac{1}{2}\)cos2x(D) cos2x |
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Answer» Option : (B) \(\frac{1}{2}\)sin2x |
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