Need to overcome erroneous result when differentiating natural log of a fractionI am trying to differentiate the following: l n ( 3 x − 8 / 6 x + 2 )my (incorrect) method is:let l n ( x ) = l n ( u )therefore when differentiating u.. l n ( u ) = 1 / u l n ( u ) = 1 / uand diff of(3x-8/6x+2) = 3/6 = 0.5so punching in values gives 0.5 ∗ ( 1 / ( 3 x − 8 / 6 x + 2 ) ) = ( 0.5 / ( 3 x − 8 / 6 x + 2 ) )Any help showing where I have gone wrong would be very much appreciated.

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Need to overcome erroneous result when differentiating natural log of a fractionI am trying to differentiate the following:  l  n  (  3  x  −  8      /    6  x  +  2  )my (incorrect) method is:let  l  n  (  x  )  =  l  n  (  u  )therefore when differentiating u..  l  n  (  u  )  =  1      /    u  l  n  (  u  )  =  1      /    uand diff of(3x-8/6x+2) = 3/6 = 0.5so punching in values gives  0.5  ∗  (  1      /    (  3  x  −  8      /    6  x  +  2  )  )  =  (  0.5      /    (  3  x  −  8      /    6  x  +  2  )  )Any help showing where I have gone wrong would be very much appreciated.

Cristal Obrien · Accepted Answer

l  n  [  (  3  x  −  8  )      /    (  6  x  +  2  )  ]  =  l  n  [  3  x  −  8  ]  −  l  n  [  6  x  +  2  ]whose derivative is      3          3      x      −      8        −      6          6      x      +      2        .

llunallenaipg5r · Accepted Answer

The derivative of an   f  (  g  (  x  )  ) function, considering that both functions are differentiable, is :       f    ′    (  g  (  x  )  )      g    ′    (  x  ) from the chain rule.What you have to find is :       d          d      x        f  (  x  )  =      d          d      x        l  n  (            3      x      −      8              6      x      +      2        )Note that :  [  l  n  (  g  (  x  )  )      ]    ′    =      1          g      (      x      )            g    ′    (  x  )Then it is :      d          d      x        f  (  x  )  =      f    ′    (  x  )  =  [  l  n  (            3      x      −      8              6      x      +      2        )      ]    ′    =            6      x      +      2              3      x      −      8        (            6      x      +      2              3      x      −      8            )    ′  You can also write   l  n  (            3      x      −      8              6      x      +      2        )  =  l  n  (  3  x  −  8  )  −  l  n  (  6  x  +  2  ) and then differentiate this expression.You can continue from then on.

Need to overcome erroneous result when differentiating natural log of a fraction I am trying to dif

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