g ( x ) = ∫ 0 2 x sin ⁡ ( t 2 ) d t , g'(0)=

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g    (    x    )    =          ∫              0                              2          x                      sin    ⁡          (              t        2            )            d      t      , g&#039;(0)=

didihkanvuc · Accepted Answer

g  (  x  )  =  F  (      2    x    )  −  F  (  0  )  .      2    x    =      e          x      ln      ⁡      2      the derivative of       2    x   is   (  ln  ⁡  2  )      e          x      ln      ⁡      2        =  (  ln  ⁡  2  )      2    x        g    ′    (  x  )  =  (  ln  ⁡  2  )      2    x        F    ′    (      2    x    )  .      F    ′    (      2    x    )  =  sin  ⁡  (  (      2    x        )    2    )  =  sin  ⁡  (      2          2      x        )      g    ′    (  x  )  =  (  ln  ⁡  2  )      2    x    sin  ⁡  (      2          2      x        )  .Let   u  =      2    x   and a=0  g  (  x  )  =      ∫    a    u    sin  ⁡  (      t    2    )    d  t  .            d      g              d      x        =            d      g              d      u                  d      g              d      x        .But             d      g              d      u        =  sin  ⁡  (      u    2    ) and             d      u              d      x        =  (  ln  ⁡  2  )      2    x  , so we are  finished.

g(x)=int ^2^x _0 sin(t^2)dt, g'(0)=

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