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anonymous
 5 years ago
Find dy/dx for: y = cos^2(sinx)
anonymous
 5 years ago
Find dy/dx for: y = cos^2(sinx)

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myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0y'=(sinx)'2*cos(sinx)(cos)'(sinx)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey is it y= cos\[(sinx)^{2}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0y'=(cosx)2[cos(sinx)](sin(sinx))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Huh? \[\frac{dy}{dx} = 2cos(sin\ x)(sin(sin\ x))(cos\ x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whole lot of chain rule goin on.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0it might be easier if you rewrite y=[cos(sinx)]^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sine sine everywhere a sine..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, it's just chain rule. First we take the derivative of \(cos^2(sin\ x)\) just as if it were \(u^2\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Those 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We get 2u*(the derivative of u)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where u is the cos(sin x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where p is sin x. So the derivative of p is the derivative of sin x. Which is cos x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm trying it right now. YOU ARE GETTING A MEDAL NO MATTER WHAT :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So dy/dx = 2(sin(sinx))(cosx)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Close, you're missing a cos(sin x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually my second post with the 'simplifications' are totally wrong for that reason ;p But the first version I gave was correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, isn't it: 2(cos(sinx))(sinx(sinx))(cosx) (I forgot the 2nd part of the chain rule >_>)
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