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dangbeau Group Title

Find the derivative f(θ)= secθ^2 How to solve this?

  • one year ago
  • one year ago

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  1. inkyvoyd Group Title
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    use the chain rule

    • one year ago
  2. inkyvoyd Group Title
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    d/dx f(g(x))=f'(g(x))g'(x)

    • one year ago
  3. andrea408 Group Title
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    f(θ) = y f'(θ) = dy/dθ u = sec(θ) du/dθ = sec(θ)tan(θ) y = u² dy/du = 2u Then, f'(θ) = 2usec(θ)tan(θ) ==> 2(sec²(θ))sec(θ)tan(θ) ==> 2sec³(θ)tan(θ) I hope this helps!

    • one year ago
  4. inkyvoyd Group Title
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    f(x)=x^2 g(th)=sec th

    • one year ago
  5. kausarsalley Group Title
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    |dw:1362873411456:dw| |dw:1362873506755:dw|

    • one year ago
  6. dangbeau Group Title
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    so the answer is the same as me f'(θ)= \[2\sec ^{2}\theta \tan \theta\], right? but i don't know why the right answer is \[2\theta \sec \theta ^{2} \tan \theta ^{2}\] can anyone explain it?

    • one year ago
  7. inkyvoyd Group Title
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    the answer is the one that you got and others gave. http://www.wolframalpha.com/input/?i=d%2Fdx+sec^2+x the answer 2th sec^2 th tan^2 th is incorrect.

    • one year ago
  8. kausarsalley Group Title
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    i also think so!

    • one year ago
  9. dangbeau Group Title
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    hm... i have another answer \[\sec ^{2} \theta=\sec \theta ^{2}\] it's equal, right?

    • one year ago
  10. dangbeau Group Title
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    i mean question

    • one year ago
  11. inkyvoyd Group Title
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    no, of course not. sec^2(th)= sec th*sec th sec(th^2)=sec(th*th)

    • one year ago
  12. NoelGreco Group Title
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    No, it's not correct.Re-write sec^2 x as (sec x)^2, then power and chain rules.

    • one year ago
  13. dangbeau Group Title
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    wait, this one http://www.wolframalpha.com/input/?i=d%2Fdx+sec^2+x you solve it as f(x)=sec^2x, but i ask f(x)=secx^2 o.0

    • one year ago
  14. inkyvoyd Group Title
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    oh - http://www.wolframalpha.com/input/?i=d%2Fdx+sec+x^2

    • one year ago
  15. dangbeau Group Title
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    how to get that answer?

    • one year ago
  16. inkyvoyd Group Title
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    use f(x)=sec x g(x)=x^2 then f'(g(x))g'(x)=tan(x^2)sec(x^2)*d/dx x^2=2x tan(x^2)sec(x^2)

    • one year ago
  17. inkyvoyd Group Title
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    does that make sense?

    • one year ago
  18. dangbeau Group Title
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    ah ok i got it

    • one year ago
  19. inkyvoyd Group Title
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    okay - so the chain rule makes more sense to you now?

    • one year ago
  20. dangbeau Group Title
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    yup, thanks sooo much ^^

    • one year ago
  21. inkyvoyd Group Title
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    You're welcome :)

    • one year ago
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