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Math Help - long chain rule

  1. #1
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    long chain rule

    my brain is not working tonight...

    what's the best way to deal with this prob using the chain rule?

    cos^2(sin x)
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  2. #2
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    Quote Originally Posted by thedoge View Post
    my brain is not working tonight...
    Ha ha ha.
    what's the best way to deal with this prob using the chain rule?

    cos^2(sin x)
    It is saying, find y' for,
    (\cos (\sin x))^2=u^2, u=\cos v, v=\sin x

    \frac{dy}{du}=2u
    \frac{du}{dv}=-\sin v
    \frac{dv}{dx}=\cos x

    Thus,
    \frac{dy}{du}\cdot \frac{du}{dv}\cdot \frac{dv}{dx}=\frac{dy}{dx}
    Thus,
    -2u\sin v\cos x=-2\cos v\sin v\cos x=-2\cos (\sin x)\sin (\sin x)\cos x
    If you are good at trignometry you can spot the double angle identity,
    -\sin (2\sin x)\cos x
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