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

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To find the derivative of y = ln((√3 - √2 cos x)/(√3 + √2 cos x), it is recommended to use the properties of logarithms to simplify the expression into two separate logarithms. This approach makes calculating the derivative easier by allowing the use of the quotient rule on simpler functions. The correct derivative can be derived as y' = (2√6 sin x)/(3 - 2 cos² x). Acknowledging the simplification helps avoid confusion and errors in the differentiation process. Utilizing logarithmic properties can significantly streamline the solution.
Lynne
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Hi,
I need help on this:

Homework Statement



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

find y'

The Attempt at a Solution


I know that answer should be: \frac{2\sqrt6 \ sinx}{3-2cos^2 x} , but can't find it:


<br /> <br /> <br /> y&#039;= ( ln \frac{\sqrt{3}-\sqrt{2} cos x}{\sqrt{3} +\sqrt{2} cos x} )&#039; = <br /> <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} )&#039; \ <br /> <br /> =\ \frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x}\<br /> <br /> (\ \frac{(\sqrt{3}-\sqrt{2} cos x)&#039; (\sqrt{3} +\sqrt{2} cos x) - (\sqrt{3}-\sqrt{2} cos x) (\sqrt{3} +\sqrt{2} cos x)&#039; }{(\sqrt{3}+\sqrt{2} cos x)^2}\ )=


= \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 })\ )<br />


<br /> <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)<br />



<br /> <br /> =\frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x} \ \frac{1}{\sqrt3}<br />
 
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You messed up when you took the derivatives while using the quotient rule.

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

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

The rest is just simplifying, which should be fun =)
 
Last edited:
Lynne said:
Hi,
I need help on this:

Homework Statement



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

find y&#039;

Actually, to make your life easier, use properties of the natural log. For example,
\ln\left( \frac{f(x)}{g(x)} \right) = \ln f(x) - \ln g(x)
Then your problem changes to finding y&#039; of
y = \ln \left( \sqrt{3} - \sqrt{2}\cos x \right) - \ln \left( \sqrt{3} + \sqrt{2}\cos x \right)
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.
 
Thank you very, very much!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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