long chain rule

**thedoge** · Jan 30th 2007, 05:03 PM

my brain is not working tonight...

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

cos^2(sin x)

**ThePerfectHacker** · Jan 30th 2007, 05:47 PM

Originally Posted by thedoge

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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