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Derivativeshard
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If f(x)=cos⁡(x)cos⁡(2x)cos⁡(4x)…cos⁡(2nx)f(x) = \cos(x) \cos(2x) \cos(4x) \dots \cos(2^n x)f(x)=cos(x)cos(2x)cos(4x)…cos(2nx), what is f′(0)f'(0)f′(0)?