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dangbeau

  • 2 years ago

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

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  1. inkyvoyd
    • 2 years ago
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    use the chain rule

  2. inkyvoyd
    • 2 years ago
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    d/dx f(g(x))=f'(g(x))g'(x)

  3. andrea408
    • 2 years ago
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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!

  4. inkyvoyd
    • 2 years ago
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    f(x)=x^2 g(th)=sec th

  5. kausarsalley
    • 2 years ago
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    |dw:1362873411456:dw| |dw:1362873506755:dw|

  6. dangbeau
    • 2 years ago
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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?

  7. inkyvoyd
    • 2 years ago
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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.

  8. kausarsalley
    • 2 years ago
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    i also think so!

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

  10. dangbeau
    • 2 years ago
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    i mean question

  11. inkyvoyd
    • 2 years ago
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    no, of course not. sec^2(th)= sec th*sec th sec(th^2)=sec(th*th)

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

  13. dangbeau
    • 2 years ago
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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

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

  15. dangbeau
    • 2 years ago
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    how to get that answer?

  16. inkyvoyd
    • 2 years ago
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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)

  17. inkyvoyd
    • 2 years ago
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    does that make sense?

  18. dangbeau
    • 2 years ago
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    ah ok i got it

  19. inkyvoyd
    • 2 years ago
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    okay - so the chain rule makes more sense to you now?

  20. dangbeau
    • 2 years ago
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    yup, thanks sooo much ^^

  21. inkyvoyd
    • 2 years ago
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    You're welcome :)

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