1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do some algebra.

    Find a common denominator for [itex]\displaystyle 1+\frac{x}{\sqrt{x^2+1}}\,.[/itex] & write as one fraction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook