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Math Help - 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 y=\ln\cos (4x) may be differentiated by applying the Chain Rule twice, knowing that

    \begin{aligned}<br />
\frac{d}{dx}\ln x&=\frac{1}{x}\\<br />
\frac{d}{dx}\cos x&=-\sin x\\<br />
\frac{d}{dx}(kx)&=k.<br />
\end{aligned}

    For reference, the Chain Rule states that

    \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!

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

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

    So...

    \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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