$\displaystyle f(x) = 2x^2sin3x $ derivative $\displaystyle f '(x) = 4xsin3x + 2x^2(3cos3x) $ but simplified its $\displaystyle f '(x) = 2x(2sin3x + 3xcos3x) $ how did they manage to simplify it like that? thank u
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$\displaystyle f '(x) = 4xsin3x + 2x^2(3cos3x) = 4xsin3x + 6x^2cos3x$ Now factor out 2x
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