Help with logarithmic differentiation problems ( 1 ) Find y ′ of y = 8 x My try: ln ⁡ y = ln ⁡ ( 8 ) x 1 y y ′ = x ln ⁡ 8I don't know how to proceed with right side. ( 2 ) Find y ′ of y = ( t + 4 ) ( t + 6 ) ( t + 7 ) .This one I have no idea what to do so I don't have any work to show. My text says to use logarithmic differentiation, but still I don't how to solve this.Thank you.

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Help with logarithmic differentiation problems      (    1    )   Find       y          ′       of   y  =      8                  x            My try:  ln  ⁡  y  =  ln  ⁡  (  8      )                  x                        1      y            y          ′        =      x    ln  ⁡  8I don&#039;t know how to proceed with right side.      (    2    )   Find       y          ′       of   y  =  (  t  +  4  )  (  t  +  6  )  (  t  +  7  )  .This one I have no idea what to do so I don&#039;t have any work to show. My text says to use logarithmic differentiation, but still I don&#039;t how to solve this.Thank you.

Esteban Johnson · Accepted Answer

(1) Note that   ln  ⁡      (          a              b              )    =  b  ln  ⁡      (    a    )   so, for   y  =      8                  x            ,   ln  ⁡  y  =      x    ln  ⁡      (    8    )  . Differentiating, we get             y      ′        y    =            ln      ⁡              (        8        )                    2              x             and so       y    ′    =  ln  ⁡      (    8    )              8                        x                            2              x            (2) I assume you&#039;re required to use logarithmic differentiation for this.Remember that   ln  ⁡      (    a    b    )    =  ln  ⁡  a  +  ln  ⁡  b. Therefore, for   y  =      (    t    +    4    )        (    t    +    6    )        (    t    +    7    )  , we have   ln  ⁡  y  =  ln  ⁡      (    t    +    4    )    +  ln  ⁡      (    t    +    6    )    +  ln  ⁡      (    t    +    7    )  . From there, differentiating, we have             y      ′        y    =      1          t      +      4        +      1          t      +      6        +      1          t      +      7      . Therefore,       y    ′    =      (    t    +    4    )        (    t    +    6    )        (    t    +    7    )        (          1              t        +        4              +          1              t        +        6              +          1              t        +        7              )  .

Peyton Velez · Accepted Answer

ln  ⁡  y  =      x    ln  ⁡  8, so       1    y              d      y              d      x        =      1          2              x              ln  ⁡  8. Therefore      y    ′    =      y          2              x              ln  ⁡  8  =            8                        x                            2              x              ln  ⁡  8For the second question, except for   t  =  −  4  ,  −  6  ,  −  7, you may take logarithm of both sides and note that       1          d      t        ln  ⁡  (  t  +  a  )  =      1          t      +      a      , and then carry on as above. However you can solve this problem by simple differentiation then you need not even take care of whether   t takes the aforementioned values or not.

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