How do I find the derivative of ln with fractions and trig functions?

  • Thread starter Thread starter Lynne
  • Start date Start date
  • Tags Tags
    Derivative
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
Lynne
Messages
12
Reaction score
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] <br /> <br /> y'= ( ln \frac{\sqrt{3}-\sqrt{2} cos x}{\sqrt{3} +\sqrt{2} cos x} )' =[/tex] [tex] <br /> \cfrac{1}{ \cfrac{\sqrt{3}-\sqrt{2} cos x} { \sqrt{3} +\sqrt{2} cos x }}\ <br /> <br /> (\frac{\sqrt{3}-\sqrt{2} cos x} {\sqrt{3} +\sqrt{2} cos x} )'[/tex] [tex]\ <br /> <br /> =\ \frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x}\<br /> <br /> (\ \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} \<br /> <br /> ( \ ( \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) }} <br /> <br /> { (\sqrt{3}+\sqrt{2} cos x)^2 })\ )[/tex]


[tex] <br /> = \frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x} <br /> (\frac{1}{2\sqrt3}-\frac{cosx}{2\sqrt2} +\sqrt2sin x +\frac{1}{2\sqrt3}+\frac{cosx}{2\sqrt2}-\sqrt2sin x)[/tex]



[tex] <br /> =\frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x} \ \frac{1}{\sqrt3}[/tex]
 
Physics news on Phys.org
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:
Lynne said:
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.
 
n!kofeyn said:
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.