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Derivative using Logarithmic differentation

  1. Jun 18, 2014 #1
    Need to find derivative using logarithmic differentiation
    [itex]y = \sqrt{x(x+1)}[/itex]

    My attempt
    [itex]ln y = ln \sqrt{x(x+1)}[/itex]

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

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

    [itex]\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}[/itex]

    [itex]\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}[/itex]

    [itex]\frac{dy}{dx}= y(\frac{1}{2x} + \frac{1}{x+1})[/itex]

    [itex]\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{1}{2x} + \frac{1}{x+1})[/itex]

    [itex]\frac{dy}{dx}= \frac{\sqrt{x(x+1)}}{2x} + \frac{\sqrt{x(x+1)}}{x+1}[/itex]

    answer in book [itex]\frac{2x+1}{2\sqrt{x(x+1))}} [/itex]
     
  2. jcsd
  3. Jun 18, 2014 #2

    verty

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    You skipped a few steps, do it more carefully and you'll see that there were mistakes.
     
  4. Jun 18, 2014 #3
    I made some corrections

    [itex]ln y = ln \sqrt{x(x+1)}[/itex]

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

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

    [itex]\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}[/itex]

    [itex]\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}[/itex]

    [itex]\frac{1}{y}= \frac{1}{x} + \frac{1}{x+1}[/itex]

    [itex]\frac{dy}{dx}= y(\frac{1}{x} + \frac{1}{x+1})[/itex]

    [itex]\frac{dy}{dx}= y(\frac{x+1+x}{x(x+1)})[/itex]

    [itex]\frac{dy}{dx}= y(\frac{2x+1}{x(x+1)})[/itex]

    [itex]\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{2x+1}{x(x+1)})[/itex]

    [itex]\frac{dy}{dx}= (\frac{\sqrt{x(x+1)}(2x+1)}{x(x+1)})[/itex]

    answer in book [itex]\frac{2x+1}{2\sqrt{x(x+1))}} [/itex]
     
  5. Jun 18, 2014 #4

    verty

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    Can you see that your answer is twice as large as the book's answer? You've lost a factor of 1/2 somewhere.
     
  6. Jun 18, 2014 #5

    Matterwave

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    Gold Member

    $$a\ln(bc)=a\ln(b)+a\ln(c)\neq a\ln(b)+\ln(c)$$

    Somewhere in your derivation you made this mistake. Hopefully you can find it.
     
  7. Jun 21, 2014 #6

    HallsofIvy

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    This is where you made your mistake.
    [itex]ln(y)= \frac{1}{2}(ln(x)+ ln(x+ 1))= \frac{1}{2}ln(x)+ \frac{1}{2}ln(x+ 1)[/itex]

     
  8. Jun 21, 2014 #7
    So, although you have found the mistake that lead to the wrong answer I still have to ask: What is going on on lines 4 and 5?
     
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