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Chain Rule Help

  1. Jun 18, 2006 #1
    [tex]y = ln[\sqrt(x^2+2)][/tex]

    I changed it to [tex] y = ln[(x^2+2)^{1/2}] [/tex]

    I know the derivative of [tex]ln = \frac {1}{x}[/tex]

    I'm not sure how to proceed.
    Last edited by a moderator: Jun 19, 2006
  2. jcsd
  3. Jun 18, 2006 #2
    set the argument of the natural log equal to u. Take derivative of u!
  4. Jun 18, 2006 #3
    If I knew how to take the derivative I wouldn't have asked!

    The way it looks to me I get:

    [tex]\frac {1}{(x^2 + 2)^{1/2}} * \frac{1}{2}(x^2 + 2)^{-1/2}[/tex]

    I'm not sure this is right though
    Last edited: Jun 18, 2006
  5. Jun 18, 2006 #4
    you have to usea certain method, perhaps a certain rule that you have been studying...
  6. Jun 18, 2006 #5
    So i'm guessing the rule in the topic is wrong then?
  7. Jun 18, 2006 #6
    what I mean is you need to do the chain rule one more time. this is an example of doing multiple chain rules in 1 problem.
  8. Jun 18, 2006 #7
    Like This?

    [tex]\frac {1}{(x^2 + 2)^{1/2}} * \frac{-1}{4}(x^2 + 2)^{-1.5} (2x)[/tex]
  9. Jun 18, 2006 #8
    May I make a suggestion to ease the diifficulty? Here's ahint

    [tex] ln(a^b) = bln(a) [/tex]

    That should deal with the square root. Then use a u substitution on the x^2 + 2 inside. That may simplify the problem a bit.

    Edit: Oops, mistyped that logarithm...
  10. Jun 18, 2006 #9
    So I can change it to this:

    [tex] y = \frac {1}{2}ln(x^2+2) [/tex]

    Then Use the Product Rule Instead?

    If so, I get:

    [tex] y' = 1x \frac {1x}{x^2 +2}[/tex]
    Last edited: Jun 18, 2006
  11. Jun 19, 2006 #10


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    Product rule? What product? If you mean the "2" times "ln...", since the derivative of a constant, normally you don't use the product rule in a situation like that: (2f(x))'= 2f'(x). But, you did get the correct answer!
    (I might be inclined not to write the 1s.)
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