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

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Homework Help Overview

The discussion revolves around finding the derivative of the function \( y = \ln \frac{\sqrt{3}-\sqrt{2} \cos x}{\sqrt{3} +\sqrt{2} \cos x} \), which involves logarithmic properties and trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of the quotient rule for differentiation and question the original poster's application of it. Some suggest using properties of logarithms to simplify the differentiation process.

Discussion Status

There are multiple approaches being discussed, with some participants offering alternative methods that may simplify the problem. Guidance has been provided regarding the application of logarithmic properties, but no consensus has been reached on the best approach.

Contextual Notes

Participants note potential mistakes in derivative calculations and the complexity of simplifying the expressions involved. The original poster expresses uncertainty about their 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!
 

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