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Math Help - Derivative with two natural logarithms: Can't figure out which rule to use

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    Junior Member ReneG's Avatar
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    Derivative with two natural logarithms: Can't figure out which rule to use

    I'm given \frac{d}{dx} \ln{(\ln{(2 - \cos {x}}))}

    I know the answer is \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.
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    Re: Derivative with two natural logarithms: Can't figure out which rule to use

    Chain rule, about 3-4 times.
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    Re: Derivative with two natural logarithms: Can't figure out which rule to use

    Quote Originally Posted by ReneG View Post
    I'm given \frac{d}{dx} \ln{(\ln{(2 - \cos {x}}))}

    I know the answer is \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.
    Suppose that f is a differentiable function.
    \frac{d}{dx} \ln{(\ln{f(x)}))}=\frac{1}{\ln(f(x))}}\cdot\frac{f  '(x)}{f(x)}
    Last edited by Plato; March 17th 2013 at 04:25 PM.
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    Re: Derivative with two natural logarithms: Can't figure out which rule to use

    Alright, so the chain rule says \frac{d}{dx} g(h(x)) = g'(h(x)) h'(x)

    Substituting in this case, would the first step be \ln'{(\ln{2 - \cos{x}})} \frac{d}{dx} \ln{(2 - \cos{x})}

    How would I find the derivative of that first outside function?
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    Re: Derivative with two natural logarithms: Can't figure out which rule to use

    Quote Originally Posted by ReneG View Post
    I'm given \frac{d}{dx} \ln{(\ln{(2 - \cos {x}}))}

    I know the answer is \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.
    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 \frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dv}{du} \cdot \frac{dy}{dv}.

    We have \displaystyle y = \ln{\left\{ \ln{ \left[ 2 - \cos{(x)} \right]} \right\} }. Let \displaystyle u = 2 - \cos{(x)} \implies y = \ln{ \left[ \ln{ ( u ) } \right] } , then let \displaystyle v = \ln{(u)} \implies y = \ln{(v)} . Then

    \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 \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] }}.
    Thanks from ReneG
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    Junior Member ReneG's Avatar
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    Re: Derivative with two natural logarithms: Can't figure out which rule to use

    You never cease making such a rigorous subject so easy to understand. I greatly appreciate it, thank you!
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    Re: Derivative with two natural logarith: Can't figure out which rule to use

    Refer attached solution.
    Attached Thumbnails Attached Thumbnails Derivative with two natural logarithms: Can't figure out which rule to use-ans.jpg  
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