# Derivative help

1. Apr 18, 2009

### Lynne

Hi,
I need help on this:

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

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

find $$y'$$

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

$$y'= ( ln \frac{\sqrt{3}-\sqrt{2} cos x}{\sqrt{3} +\sqrt{2} cos x} )' =$$ $$\cfrac{1}{ \cfrac{\sqrt{3}-\sqrt{2} cos x} { \sqrt{3} +\sqrt{2} cos x }}\ (\frac{\sqrt{3}-\sqrt{2} cos x} {\sqrt{3} +\sqrt{2} cos x} )'$$ $$\ =\ \frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x}\ (\ \frac{(\sqrt{3}-\sqrt{2} cos x)' (\sqrt{3} +\sqrt{2} cos x) - (\sqrt{3}-\sqrt{2} cos x) (\sqrt{3} +\sqrt{2} cos x)' }{(\sqrt{3}+\sqrt{2} cos x)^2}\ )=$$

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

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

$$=\frac{\sqrt{3}+\sqrt{2} cos x}{\sqrt{3} -\sqrt{2} cos x} \ \frac{1}{\sqrt3}$$

2. Apr 18, 2009

### Dunkle

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: Apr 18, 2009
3. Apr 18, 2009

### n!kofeyn

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 $y'$ 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.

4. Apr 18, 2009

### Dunkle

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. Apr 18, 2009

### Lynne

Thank you very, very much!