anonymous
  • anonymous
Find dy/dx for: y = cos^2(sinx)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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myininaya
  • myininaya
y'=(sinx)'2*cos(sinx)(cos)'(sinx)
anonymous
  • anonymous
hey is it y= cos\[(sinx)^{2}\]
myininaya
  • myininaya
y'=(cosx)2[cos(sinx)](-sin(sinx))

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anonymous
  • anonymous
Huh? \[\frac{dy}{dx} = 2cos(sin\ x)(-sin(sin\ x))(cos\ x)\]
anonymous
  • anonymous
Whole lot of chain rule goin on.
myininaya
  • myininaya
it might be easier if you rewrite y=[cos(sinx)]^2
anonymous
  • anonymous
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.
anonymous
  • anonymous
Err sines.
anonymous
  • anonymous
Sine sine everywhere a sine..
anonymous
  • anonymous
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?
anonymous
  • anonymous
No, it's just chain rule. First we take the derivative of \(cos^2(sin\ x)\) just as if it were \(u^2\)
anonymous
  • anonymous
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?
anonymous
  • anonymous
We get 2u*(the derivative of u)
anonymous
  • anonymous
Where u is the cos(sin x)
anonymous
  • anonymous
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)
anonymous
  • anonymous
Where p is sin x. So the derivative of p is the derivative of sin x. Which is cos x.
anonymous
  • anonymous
Does that make sense?
anonymous
  • anonymous
I'm trying it right now. YOU ARE GETTING A MEDAL NO MATTER WHAT :D
anonymous
  • anonymous
So dy/dx = 2(-sin(sinx))(cosx)?
anonymous
  • anonymous
Close, you're missing a cos(sin x)
anonymous
  • anonymous
Actually my second post with the 'simplifications' are totally wrong for that reason ;p But the first version I gave was correct.
anonymous
  • anonymous
Wait, isn't it: 2(cos(sinx))(-sinx(sinx))(cosx) (I forgot the 2nd part of the chain rule >_>)
anonymous
  • anonymous
and yes it is :D

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