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anonymous

  • 5 years ago

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

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  1. myininaya
    • 5 years ago
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    y'=(sinx)'2*cos(sinx)(cos)'(sinx)

  2. anonymous
    • 5 years ago
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    hey is it y= cos\[(sinx)^{2}\]

  3. myininaya
    • 5 years ago
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    y'=(cosx)2[cos(sinx)](-sin(sinx))

  4. anonymous
    • 5 years ago
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    Huh? \[\frac{dy}{dx} = 2cos(sin\ x)(-sin(sin\ x))(cos\ x)\]

  5. anonymous
    • 5 years ago
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    Whole lot of chain rule goin on.

  6. myininaya
    • 5 years ago
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    it might be easier if you rewrite y=[cos(sinx)]^2

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    Err sines.

  9. anonymous
    • 5 years ago
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    Sine sine everywhere a sine..

  10. anonymous
    • 5 years ago
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    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?

  11. anonymous
    • 5 years ago
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    No, it's just chain rule. First we take the derivative of \(cos^2(sin\ x)\) just as if it were \(u^2\)

  12. anonymous
    • 5 years ago
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    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?

  13. anonymous
    • 5 years ago
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    We get 2u*(the derivative of u)

  14. anonymous
    • 5 years ago
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    Where u is the cos(sin x)

  15. anonymous
    • 5 years ago
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    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)

  16. anonymous
    • 5 years ago
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    Where p is sin x. So the derivative of p is the derivative of sin x. Which is cos x.

  17. anonymous
    • 5 years ago
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    Does that make sense?

  18. anonymous
    • 5 years ago
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    I'm trying it right now. YOU ARE GETTING A MEDAL NO MATTER WHAT :D

  19. anonymous
    • 5 years ago
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    So dy/dx = 2(-sin(sinx))(cosx)?

  20. anonymous
    • 5 years ago
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    Close, you're missing a cos(sin x)

  21. anonymous
    • 5 years ago
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    Actually my second post with the 'simplifications' are totally wrong for that reason ;p But the first version I gave was correct.

  22. anonymous
    • 5 years ago
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    Wait, isn't it: 2(cos(sinx))(-sinx(sinx))(cosx) (I forgot the 2nd part of the chain rule >_>)

  23. anonymous
    • 5 years ago
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    and yes it is :D

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