Find dy/dx for: y = cos^2(sinx)

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Find dy/dx for: y = cos^2(sinx)

Mathematics
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y'=(sinx)'2*cos(sinx)(cos)'(sinx)
hey is it y= cos\[(sinx)^{2}\]
y'=(cosx)2[cos(sinx)](-sin(sinx))

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Huh? \[\frac{dy}{dx} = 2cos(sin\ x)(-sin(sin\ x))(cos\ x)\]
Whole lot of chain rule goin on.
it might be easier if you rewrite y=[cos(sinx)]^2
Can simplify to \[-2cos^2(sin\ x)*sin(sin\ x)\] or \[2(sin^3(sin\ x)) - sin(sin\ x))\] if you prefer only dealing with signs.
Err sines.
Sine sine everywhere a sine..
ok so those are simplified versions. So to find the derivative of it, you product rule the first one (somehow). And the second one, you do product rule for the two terms?
No, it's just chain rule. First we take the derivative of \(cos^2(sin\ x)\) just as if it were \(u^2\)
Those equations look crazy confusing. I have this question on my test tomorrow, and theres about a 2 inches of blank space to find it. Is there an easier way?
We get 2u*(the derivative of u)
Where u is the cos(sin x)
So then to find the derivative of u we take the derivative of the cos(sin x) just as if it were cos p. And we get -sin(p)*(derivative of p)
Where p is sin x. So the derivative of p is the derivative of sin x. Which is cos x.
Does that make sense?
I'm trying it right now. YOU ARE GETTING A MEDAL NO MATTER WHAT :D
So dy/dx = 2(-sin(sinx))(cosx)?
Close, you're missing a cos(sin x)
Actually my second post with the 'simplifications' are totally wrong for that reason ;p But the first version I gave was correct.
Wait, isn't it: 2(cos(sinx))(-sinx(sinx))(cosx) (I forgot the 2nd part of the chain rule >_>)
and yes it is :D

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