y = ( 3 x 2 + 2 ) l n x The answer to this is: d y d x = ( 3 x 2 + 2 ) l n x ( 1 x l n ( 3 x 2 + 2 ) + 6 x l n x 3 x 2 + 2 )What I'm coming up with is: d y d x = ( 3 x 2 + 2 ) l n x ( 1 x l n ( 3 x 2 + 2 ) + 6 x 3 x 2 + 2 )What I'm not understanding is where the 6 x l n x 3 x 2 + 2 comes from, if anyone could explain this I'd really appreciate it.

Question

y  =  (  3      x          2        +  2      )          l      n      x      The answer to this is:            d      y              d      x        =  (  3      x          2        +  2      )          l      n      x        (      1    x    l  n  (  3      x          2        +  2  )  +            6      x      l      n      x              3              x                  2                    +      2        )What I&#039;m coming up with is:            d      y              d      x        =  (  3      x          2        +  2      )          l      n      x        (      1    x    l  n  (  3      x          2        +  2  )  +            6      x              3              x                  2                    +      2        )What I&#039;m not understanding is where the             6      x      l      n      x              3              x                  2                    +      2       comes from, if anyone could explain this I&#039;d really appreciate it.

umgangistbf · Accepted Answer

Using logarithmic differentiation, that is,  ln  ⁡  (  y  )  =  ln  ⁡  (  x  )  ln  ⁡  (  3      x    2    +  2  )  .Now, when you take the derivative, the left-hand side becomes            y      ′        y    .Use the product rule on the right-hand side, and don&#039;t forget   y  =  (  3      x    2    +  2      )          ln      ⁡      (      x      )        . Thus, altogether you&#039;ll get            y      ′        y    =      1    x    ln  ⁡  (  3      x    2    +  2  )  +            ln      ⁡      (      x      )              3              x        2            +      2        (  6  x  )  .Above, the 6x comes from the chain rule. Multiply both sides by   y  =  (  3      x    2    +  1      )          ln      ⁡      (      x      )       and we get      y    ′    =  (  3      x    2    +  2      )          ln      ⁡      (      x      )            (          1      x        ln    ⁡    (    3          x      2        +    2    )    +                  6        x        ln        ⁡        (        x        )                    3                  x          2                +        2              )    .We can perhaps go a little further in simplification to get      y    ′    =  (  3      x    2    +  2      )          ln      ⁡      (      x      )      +      1            (          1      x        +                  6        x        ln        ⁡        (        x        )                    (        3                  x          2                +        2                  )          2                      )    .Here, I have factored a   (  3      x    2    +  2  ) from each term, thus the second term will get a square in the denominator, and the factor in front of the fraction gets an extra +1 in the numerator.

I've been working through my practice problems and came across one that has stumped me. y=(3x^2+2)^(ln x) What I'm not understanding is where the (6x ln x)/(3x^2+2) comes from, if anyone could explain this.

Answered question

Answer & Explanation

New Questions in Calculus 1