I'm given $\displaystyle \frac{d}{dx} \ln{(\ln{(2 - \cos {x}}))}$
I know the answer is $\displaystyle \frac{\sin{x}}{(2 - \cos{x})(\ln{2 - \cos{x}})}$
but have no idea which rule I should use, and what I should do to carry it out.
I'm given $\displaystyle \frac{d}{dx} \ln{(\ln{(2 - \cos {x}}))}$
I know the answer is $\displaystyle \frac{\sin{x}}{(2 - \cos{x})(\ln{2 - \cos{x}})}$
but have no idea which rule I should use, and what I should do to carry it out.
Alright, so the chain rule says $\displaystyle \frac{d}{dx} g(h(x)) = g'(h(x)) h'(x)$
Substituting in this case, would the first step be $\displaystyle \ln'{(\ln{2 - \cos{x}})} \frac{d}{dx} \ln{(2 - \cos{x})}$
How would I find the derivative of that first outside function?
You have the function 2 - cos(x) inside a logarithm function, inside another logarithm function. Since this composition is three functions deep, you will need three links in your chain when you use the chain rule, i.e. $\displaystyle \displaystyle \frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dv}{du} \cdot \frac{dy}{dv}$.
We have $\displaystyle \displaystyle y = \ln{\left\{ \ln{ \left[ 2 - \cos{(x)} \right]} \right\} }$. Let $\displaystyle \displaystyle u = 2 - \cos{(x)} \implies y = \ln{ \left[ \ln{ ( u ) } \right] } $, then let $\displaystyle \displaystyle v = \ln{(u)} \implies y = \ln{(v)} $. Then
$\displaystyle \displaystyle \begin{align*} \frac{du}{dx} &= \sin{(x)} \\ \\ \frac{dv}{du} &= \frac{1}{u} \\ &= \frac{1}{2 - \cos{(x)}} \\ \\ \frac{dy}{dv} &= \frac{1}{v} \\ &= \frac{1}{\ln{(u)}} \\ &= \frac{1}{\ln{ \left[ 2 - \cos{(x)} \right] }} \end{align*}$
So $\displaystyle \displaystyle \frac{dy}{dx} = \sin{(x)} \cdot \frac{1}{2 - \cos{(x)}} \cdot \frac{1}{\ln{\left[ 2 - \cos{(x)} \right]}} = \frac{\sin{(x)}}{\left[ 2 - \cos{(x)} \right] \ln{ \left[ 2 - \cos{(x)} \right] }}$.