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Math Help - Overlapping Chain Rule?

  1. #1
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    Overlapping Chain Rule?

    Hi,

    I'm having some trouble with this problem, I'm supposed to find the derivative:

    y = \sin^2(\cos(3x^{-2}))

    I know that the Chain Rule can be applied, but to what extent? There are three terms. I tried to simplify it to

    2\cos(\cos(3x^{-2}) \cdot -\sin(3x^{-2})

    Would the next step be to differentiate the right side? Also, on the left side, you could differentiate again and again only to still have the \cos(3x^{-2}) - how do you know for certain that once is enough?

    Thanks in advance!
    Last edited by Biff; November 13th 2012 at 09:59 AM.
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  2. #2
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    Re: Overlapping Chain Rule?

    Quote Originally Posted by Biff View Post
    find the derivative:
    y = \sin^2(\cos(3x^{-2}))
    Try
    \left\{ {2\sin \left( {\cos \left( {3x^{ - 2} } \right)} \right)} \right\}\left[ {\cos \left( {\cos \left( {3x^{ - 2} } \right)} \right)} \right]\left[ { - \sin \left( {3x^{ - 2} } \right)} \right]\left[ { - 6x^{ - 3} } \right]
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  3. #3
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    Re: Overlapping Chain Rule?

    You can apply the chain rule as many times as you have strength for! Here, you have y(x)= sin^2(cos(3x^{-2}))
    Let u= 3x^{-2}. Then y(u)= sin^2(cos(u)). Let v= cos(u). Then y(v)= sin^2(v). Let w= sin(v). Then y(w)= w^2.
    Then \frac{dy}{dx}= \frac{dy}{dw}\frac{dw}{dv}\frac{dv}{du}\frac{du}{d  x} which gives what Plato said.
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  4. #4
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    Re: Overlapping Chain Rule?

    Now this makes sense - thanks guys!

    Here's how I solved it (separating it out helped):

    \frac{dy}{dx} = 2\sin(\cos(3x^{-2})) \cdot \cos(\cos(3x^{-2})) \cdot (-\sin(3x^{-2})) \cdot -6x^{-3}

    \frac{dy}{dx} = 12x^{-3}\sin(\cos(3x^{-2})) \cdot \cos(\cos(3x^{-2})) \cdot \sin(3x^{-2}))
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