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Thread: Differentiating functions with natural logs

  1. #1
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    Differentiating functions with natural logs

    Can anyone explain to me how to do this problem?


    y=ln(cos(4x))
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  2. #2
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    Welcome to the Math Help Forum!

    The function $\displaystyle y=\ln\cos (4x)$ may be differentiated by applying the Chain Rule twice, knowing that

    $\displaystyle \begin{aligned}
    \frac{d}{dx}\ln x&=\frac{1}{x}\\
    \frac{d}{dx}\cos x&=-\sin x\\
    \frac{d}{dx}(kx)&=k.
    \end{aligned}$

    For reference, the Chain Rule states that

    $\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))g'(x).$
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by iheartphysics01 View Post
    Can anyone explain to me how to do this problem?


    y=ln(cos(4x))
    Sure!

    $\displaystyle \frac{d}{dx}[\ln{u}]=\frac{u'}{u}$

    In your problem, let $\displaystyle u=\cos(4x)$. This implies $\displaystyle u'=-4\sin{(4x)}$

    So...

    $\displaystyle \frac{d}{dx}[\ln{u}]=\frac{u'}{u}=\frac{-4\sin(4x)}{\cos(4x)}=-4\tan(4x)$
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  4. #4
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    ok thanks so much!
    i will probably be using this forum a lot!
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