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Find dy/dx of y= the square root of ln x

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Find dy/dx when y=[itex]\sqrt{ln x}[/itex]


    2. Relevant equations
    d/dx of ln x is equal to 1/x times d/dx of x.


    3. The attempt at a solution
    I tried to raise the ln x to the 1/2 power instead of keeping it under a square root sign, but I had no luck. I'm struggling with Calculus. I would very much appreciate some help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2012 #2

    sharks

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    Gold Member

    To visualize the problem in an easier way, let u = ln x. Then proceed. First, differentiate the power, then the term/s under the square root.
     
  4. Sep 29, 2012 #3
    This strategy for solving the problem won't work because (ln(x))^a ≠ln(x^a) = a*ln(x). As suggested by sharks, u substitution is the best strategy here.
     
  5. Sep 30, 2012 #4
    Using u-sub:


    [itex]\frac{dy}{dx}[/itex][itex]\sqrt{lnx}[/itex] | u=lnx

    =[itex]\sqrt{u}[/itex]'*u'

    =[itex]\frac{1}{2\sqrt{lnx}}[/itex]*[itex]\frac{1}{x}[/itex]

    =[itex]\frac{1}{2x\sqrt{lnx}}[/itex]

    It's been a while since I've done derivatives, but I think this should be the correct way of working it.
     
  6. Sep 30, 2012 #5

    dextercioby

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    Science Advisor
    Homework Helper

    There's a nice trick at hand.

    [tex] y^2 (x) = \ln x [/tex]

    Differentiate both sides with respect to x.
     
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