logarithmic derivative of x e ( x 2 + cos ⁡ x ) I'm having a hard time taking the derivative of f ( x ) = x e ( x 2 + c o s x ) .I'm aware that I have to take the logarithm of both sides. ln ⁡ ( y ) = ln ⁡ ( x e ( x 2 + cos ⁡ x ) ) = ln ⁡ ( x ) ⋅ e ( x 2 + cos ⁡ x ) Which I tried to untie, so lets start: First I use the product rule: 1 y ⋅ y ′ = 1 x ⋅ e ( x 2 + cos ⁡ x ) + ln ⁡ ( x ) ⋅ e ( x 2 + cos ⁡ x ) Next the power rule: 1 y ⋅ y ′ = 1 x ⋅ e ( x 2 + cos ⁡ x ) + ln ⁡ ( x ) ∗ e ( x 2 + c o s x ) ⋅ x 2 + cos ⁡ x ⋅ e ( x 2 + cos ⁡ x ) − 1 Then I bring the y to the right: y ′ = y ⋅ ( 1 x ⋅ e ( x 2 + cos ⁡ x ) + ln ⁡ ( x ) ⋅ e ( x 2 + cos ⁡ x ) ⋅ x 2 + cos ⁡ x ⋅ e ( x 2 + cos ⁡ x ) − 1 ⋅ 2 x − sin ⁡ x ) And exchange y with the term: y ′ = x e ( x 2 + cos ⁡ x ) ⋅ ( 1 x ⋅ e ( x 2 + cos ⁡ x ) + ln ⁡ ( x ) ⋅ e ( x 2 + cos ⁡ x ) ⋅ x 2 + cos ⁡ x ⋅ e ( x 2 + cos ⁡ x ) − 1 ⋅ 2 x − sin ⁡ x ) This is extremely overwhelming for me and I have absolutely no clue if this is right, I looked in to the result of wolfram and It doesn't seem to be correct. Any help would be appreciated.

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logarithmic derivative of       x                  e                  (                      x            2                    +          cos          ⁡          x          )                    I&#039;m having a hard time taking the derivative of  f  (  x  )  =      x                  e                  (                      x            2                    +          c          o          s          x          )                      .I&#039;m aware that I have to take the logarithm of both sides.  ln  ⁡  (  y  )  =  ln  ⁡  (            x                        e                      (                          x              2                        +            cos            ⁡            x            )                                )  =  ln  ⁡  (  x  )  ⋅      e          (              x        2            +      cos      ⁡      x      )      Which I tried to untie, so lets start: First I use the product rule:      1    y    ⋅      y    ′    =      1    x    ⋅      e          (              x        2            +      cos      ⁡      x      )        +  ln  ⁡  (  x  )  ⋅      e          (              x        2            +      cos      ⁡      x      )      Next the power rule:      1    y    ⋅      y    ′    =      1    x    ⋅      e          (              x        2            +      cos      ⁡      x      )        +  ln  ⁡  (  x  )  ∗      e          (              x        2            +      c      o      s      x      )        ⋅      x    2    +  cos  ⁡  x  ⋅      e          (              x        2            +      cos      ⁡      x      )      −      1      Then I bring the y to the right:      y    ′    =  y  ⋅      (          1      x        ⋅          e              (                  x          2                +        cos        ⁡        x        )              +    ln    ⁡    (    x    )    ⋅          e              (                  x          2                +        cos        ⁡        x        )              ⋅          x      2        +    cos    ⁡    x    ⋅          e              (                  x          2                +        cos        ⁡        x        )        −        1              ⋅    2    x    −    sin    ⁡    x    )  And exchange y with the term:      y    ′    =      x                  e                  (                      x            2                    +          cos          ⁡          x          )                      ⋅      (          1      x        ⋅          e              (                  x          2                +        cos        ⁡        x        )              +    ln    ⁡    (    x    )    ⋅          e              (                  x          2                +        cos        ⁡        x        )              ⋅          x      2        +    cos    ⁡    x    ⋅          e              (                  x          2                +        cos        ⁡        x        )        −        1              ⋅    2    x    −    sin    ⁡    x    )  This is extremely overwhelming for me and I have absolutely no clue if this is right, I looked in to the result of wolfram and It doesn&#039;t seem to be correct. Any help would be appreciated.

snowbargerpgl1f · Accepted Answer

You&#039;re right until writing  ln  ⁡  f  (  x  )  =  ln  ⁡  x  ⋅      e                  x        2            +      cos      ⁡      x        .The rest, I&#039;m afraid, is completely wrong.You can differentiate both sides using the product rule:                    f        ′            (      x      )              f      (      x      )        =  (  ln  ⁡  x      )    ′    ⋅      e                  x        2            +      cos      ⁡      x        +  ln  ⁡  x  ⋅  (      e                  x        2            +      cos      ⁡      x            )    ′  Let&#039;s split the problem, now: what&#039;s   (      e                  x        2            +      cos      ⁡      x            )    ′  ? Here you can apply the chain rule:  (      e                  x        2            +      cos      ⁡      x            )    ′    =  (  2  x  −  sin  ⁡  x  )      e                  x        2            +      cos      ⁡      x        .Therefore you have                    f        ′            (      x      )              f      (      x      )        =      1    x        e                  x        2            +      cos      ⁡      x        +  ln  ⁡  x  ⋅  (  2  x  −  sin  ⁡  x  )      e                  x        2            +      cos      ⁡      x

Schenone2pare · Accepted Answer

You just need a way to keep track of what you&#039;ve done so far, and apply the chain rule in steps. Let&#039;s start where you did:  ln  ⁡  (  y  )  =  ln  ⁡  (  x  )      e                  x                  2                    +      cos      ⁡      x      Then the product rule and chain rule give                                          y            ′                    y                                    =                  [                      d                          d              x                                ln          ⁡          x          ]                          e                                    x                              2                                      +            cos            ⁡            x                          +        ln        ⁡        (        x        )                  [                      d                          d              x                                            e                                          x                                  2                                            +              cos              ⁡              (              x              )                                ]                                                  =                  1          x                          e                                    x                              2                                      +            cos            ⁡            x                          +        ln        ⁡        (        x        )                  e                                    x                              2                                      +            cos            ⁡            x                                    d                      d            x                          (                  x                      2                          +        cos        ⁡        x        )                                          =                  1          x                          e                                    x                              2                                      +            cos            ⁡            x                          +        ln        ⁡        (        x        )                  e                                    x                              2                                      +            cos            ⁡            x                          (        2        x        −        sin        ⁡        x        )        .

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