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Derivative help

  1. Apr 18, 2009 #1
    I need help on this:

    1. The problem statement, all variables and given/known data

    [tex] y = ln \frac{\sqrt{3}-\sqrt{2} cos x}{\sqrt{3} +\sqrt{2} cos x}[/tex]

    find [tex]y'[/tex]

    3. The attempt at a solution
    I know that answer should be: [tex] \frac{2\sqrt6 \ sinx}{3-2cos^2 x} [/tex] , but can't find it:


    y'= ( ln \frac{\sqrt{3}-\sqrt{2} cos x}{\sqrt{3} +\sqrt{2} cos x} )' =[/tex] [tex]

    \cfrac{1}{ \cfrac{\sqrt{3}-\sqrt{2} cos x} { \sqrt{3} +\sqrt{2} cos x }}\

    (\frac{\sqrt{3}-\sqrt{2} cos x} {\sqrt{3} +\sqrt{2} cos x} )' [/tex] [tex] \

    =\ \frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x}\

    (\ \frac{(\sqrt{3}-\sqrt{2} cos x)' (\sqrt{3} +\sqrt{2} cos x) - (\sqrt{3}-\sqrt{2} cos x) (\sqrt{3} +\sqrt{2} cos x)' }{(\sqrt{3}+\sqrt{2} cos x)^2}\ )=[/tex]

    [tex]= \frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x} \

    ( \ ( \dfrac{{\dfrac{1}{2\sqrt3}- \dfrac{cosx}{2\sqrt2}+\sqrt{2}sinx)(\sqrt{3}+\sqrt{2} cos x)-(\sqrt{3}-\sqrt{2} cos x) \ (\dfrac{1}{2\sqrt3}+ \dfrac{cosx}{2\sqrt2}-\sqrt{2}sinx) }}

    { (\sqrt{3}+\sqrt{2} cos x)^2 })\ )


    = \frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x}
    (\frac{1}{2\sqrt3}-\frac{cosx}{2\sqrt2} +\sqrt2sin x +\frac{1}{2\sqrt3}+\frac{cosx}{2\sqrt2}-\sqrt2sin x)


    =\frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x} \ \frac{1}{\sqrt3}
  2. jcsd
  3. Apr 18, 2009 #2
    You messed up when you took the derivatives while using the quotient rule.

    [tex]\frac{d}{dx}[\sqrt{3}-\sqrt{2}cos(x)] = 0-\sqrt{2}[-sin(x)] = \sqrt{2}sin(x)[/tex]

    [tex]\frac{d}{dx}[\sqrt{3}+\sqrt{2}cos(x)] = -\sqrt{2}sin(x)[/tex]

    The rest is just simplifying, which should be fun =)
    Last edited: Apr 18, 2009
  4. Apr 18, 2009 #3
    Actually, to make your life easier, use properties of the natural log. For example,
    [tex] \ln\left( \frac{f(x)}{g(x)} \right) = \ln f(x) - \ln g(x) [/tex]
    Then your problem changes to finding [itex]y'[/itex] of
    [tex] y = \ln \left( \sqrt{3} - \sqrt{2}\cos x \right) - \ln \left( \sqrt{3} + \sqrt{2}\cos x \right) [/tex]
    Computing and simplifying this derivative will be much easier.
  5. Apr 18, 2009 #4
    Yes, that is much easier! I was being so narrow-minded when approaching this problem because I was trying to figure out where Lynne made a mistake, so I used the same method.
  6. Apr 18, 2009 #5
    Thank you very, very much!
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