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Math Help - simplified derivative?

  1. #1
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    simplified derivative?

     <br /> <br />
f(x) = 2x^2sin3x<br /> <br />

    derivative

     <br /> <br />
f '(x) = 4xsin3x + 2x^2(3cos3x)<br /> <br />

    but simplified its

     <br /> <br />
f '(x) = 2x(2sin3x + 3xcos3x)<br /> <br />

    how did they manage to simplify it like that? thank u
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  2. #2
    Super Member 11rdc11's Avatar
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    f '(x) = 4xsin3x + 2x^2(3cos3x) = 4xsin3x + 6x^2cos3x

    Now factor out 2x
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