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what's the best way to deal with this prob using the chain rule?
cos^2(sin x)
Ha ha ha.Originally Posted by thedoge
It is saying, find $\displaystyle y'$ for,what's the best way to deal with this prob using the chain rule?
cos^2(sin x)
$\displaystyle (\cos (\sin x))^2=u^2$, $\displaystyle u=\cos v$, $\displaystyle v=\sin x$
$\displaystyle \frac{dy}{du}=2u$
$\displaystyle \frac{du}{dv}=-\sin v$
$\displaystyle \frac{dv}{dx}=\cos x$
Thus,
$\displaystyle \frac{dy}{du}\cdot \frac{du}{dv}\cdot \frac{dv}{dx}=\frac{dy}{dx}$
Thus,
$\displaystyle -2u\sin v\cos x=-2\cos v\sin v\cos x=-2\cos (\sin x)\sin (\sin x)\cos x$
If you are good at trignometry you can spot the double angle identity,
$\displaystyle -\sin (2\sin x)\cos x$
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