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Homework Help: Another Iffy Derivative

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]f(x) = ln(x+\sqrt{x^2+1})[/itex]

    2. Relevant equations



    3. The attempt at a solution

    First, I applied the chain rule.

    [itex][\frac{1}{x+\sqrt{x^2+1}}]Dx[x+\sqrt{x^2+1}][/itex]

    Second, to find [itex]Dx[x+\sqrt{x^2+1}][/itex], I broke it into two derivatives. Derivative of x is 1, so

    [itex]1 + Dx[\sqrt{x^2+1}][/itex]

    To find [itex]Dx[\sqrt{x^2+1}][/itex], I applied the chain rule once more.

    [itex][\frac{1}{2}][2x]\frac{1}{\sqrt{x^2+1}}[/itex]

    I simplified this result to:

    [itex]\frac{x}{\sqrt{x^2+1}}[/itex]


    Leading to and end-derivative of:

    [itex][\frac{1}{x+\sqrt{x^2+1}}][1+\frac{x}{\sqrt{x^2+1}}][/itex]

    The book gives a much cleaner answer of [itex]\frac{1}{\sqrt{x^2+1}}[/itex]

    Is my answer equivalent? If yes, how would I get to that? If no, what part of the calculus did I screw up?


    WOW, Nevermind!
     
    Last edited: Jul 27, 2011
  2. jcsd
  3. Jul 27, 2011 #2

    SammyS

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    Do some algebra.

    Find a common denominator for [itex]\displaystyle 1+\frac{x}{\sqrt{x^2+1}}\,.[/itex] & write as one fraction.
     
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