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Math Help - Studyin for final tommorow (chain rule/derivative)

  1. #1
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    Studyin for final tommorow (chain rule/derivative)

    i m pretty good on finding most derivatives.. but i m stuck on this one:

    y = cos(cos(cosx))

    can some1 show me how to find the derivative to this? i know the trick is using the chain rule. please simplyify the steps cause i sumtimes dont understand.
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  2. #2
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    y=\cos (\cos (\cos x))\implies y'=-\sin (\cos (\cos x))\cdot \left( \cos (\cos x) \right)'.

    Proceed from there, it's just a matter of chain rule.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    y=\cos (\cos (\cos x))\implies y'=-\sin (\cos (\cos x))\cdot \left( \cos (\cos x) \right)'.

    Proceed from there, it's just a matter of chain rule.
    hm not quite understanding
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Legendsn3verdie View Post
    hm not quite understanding
    The extended chain rule states the following
    \frac{d}{dx}\bigg[f(g(h(x)))\bigg]=f'(g(h(x)))\cdot{g'(h(x))}\cdot{h'(x)}
    So seeing that f(x),g(x),h(x) all equal cos(x)


    We can see that \frac{d}{dx}\bigg[\cos(\cos(\cos(x)))\bigg]=-\sin(\cos(\cos(x)))\cdot{-\sin(\cos(x))}\cdot{-\sin(x)}

    Does this make sense?

    If you are trying to practice the chain rule try these on for size

    \text{Find}\text{ }\frac{d}{dx}\bigg[\ln(\cos(x^2))\bigg]

    \text{Find}\text{ }\frac{d}{dx}\bigg[\cos(\sqrt[3]{1-\tan(x)})\bigg]

    and finally

    \text{Find}\text{ }\frac{d}{dx}\bigg[\ln\bigg(\frac{5x}{\cos(arcsin(x))}\bigg)\bigg]
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  5. #5
    Moo
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    Hello,

    I'm not sure showing him the direct formula would help lol

    Quote Originally Posted by Legendsn3verdie View Post
    hm not quite understanding
    Ok, you should know the chain rule : (u(v(x))'=v'(x) \cdot u'(v(x))

    Here, u(t)=\cos t ( u'(t)=-\sin t) and v(x)=\cos(\cos(x))=t

    \cos(\cos(\cos(x)))=u(v(x))

    -> The derivative of \cos(\cos(\cos(x))) is v'(x) \cdot u'(v(x))=\bigg(\cos(\cos(x))\bigg)' \cdot \bigg(- \sin (\cos(\cos(x)))\bigg)


    Does it make sense to you ? :x
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