Differentiate x 2 ln ⁡ ( 1 + x 2 )I simplified it till - 2 x 2 ln ⁡ ( x )Then I differentiate - 2 x 2 ( 1 x ( d d x x ) + ln ⁡ x ( 4 x ) 2 x 2 x + 4 x ln ⁡ xWhere have I gone wrong ? Because the answer I am trying to achieve is 2 x 3 1 + x 2 + 2 x ln ⁡ ( 1 + x 2 )

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Differentiate      x    2    ln  ⁡  (  1  +      x    2    )I simplified it till -  2      x    2    ln  ⁡  (  x  )Then I differentiate -  2      x    2    (            1      x        (            d              d        x              x  )  +  ln  ⁡  x  (  4  x  )                    2                  x          2                    x        +  4  x  ln  ⁡  xWhere have I gone wrong ? Because the answer I am trying to achieve is                    2                  x          3                            1        +                  x          2                      +  2  x  ln  ⁡  (  1  +      x    2    )

micelarnyiz · Accepted Answer

Simply apply product rule :            d      (      f      (      x      )      ⋅      g      (      x      )      )              d      x        =      f    ′    (  x  )  ⋅  g  (  x  )  +  f  (  x  )  ⋅      g    ′    (  x  )

nascarchic839e9 · Accepted Answer

Your simplification is wrong: you can&#039;t do that with logarithms. Directly with the product rule:            (              x        2            log      ⁡      (      1      +              x        2            )      )        ′    =  2  x  log  ⁡  (  1  +      x    2    )  +            2              x        3                    1      +              x        2

Differentiate x^2 ln(1+x^2)

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