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Derivative help

  • Thread starter Lynne
  • Start date
  • #1
12
0
Hi,
I need help on this:

Homework Statement



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

find [tex]y'[/tex]

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:


[tex]


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 })\ )
[/tex]


[tex]

= \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)
[/tex]



[tex]

=\frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x} \ \frac{1}{\sqrt3}
[/tex]
 

Answers and Replies

  • #2
56
0
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:
  • #3
537
3
Hi,
I need help on this:

Homework Statement



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

find [tex]y'[/tex]
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.
 
  • #4
56
0
Actually, to make your life easier, use properties of the natural log.
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.
 
  • #5
12
0
Thank you very, very much!
 

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