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v.s Group Title

Differentiate. f(θ) = sec θ/3 + sec θ

  • 2 years ago
  • 2 years ago

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  1. Coolsector Group Title
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    \[f(t) = \frac{ \sec(t) }{ 3 } + \sec(t)\]

    • 2 years ago
  2. Coolsector Group Title
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    this is the question ?

    • 2 years ago
  3. Coolsector Group Title
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    or it's 3+sec(t) in the denominator ?

    • 2 years ago
  4. v.s Group Title
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    yeaaa

    • 2 years ago
  5. v.s Group Title
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    in the denominator

    • 2 years ago
  6. Coolsector Group Title
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    \[f(t)=\frac{ \sec(t) }{ 3 + \sec(t) }\]

    • 2 years ago
  7. Coolsector Group Title
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    like this ?

    • 2 years ago
  8. v.s Group Title
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    yes

    • 2 years ago
  9. micahwood50 Group Title
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    In this case, use quotient rule.

    • 2 years ago
  10. Coolsector Group Title
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    grr mistake :| all over again

    • 2 years ago
  11. Coolsector Group Title
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    \[f'(t) = \frac{ \sec(t)' * (3+\sec(t)) - (3+\sec(t))' * \sec(t) }{ (3+\sec(t))^2 }\]

    • 2 years ago
  12. Coolsector Group Title
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    \[f'(t) = \frac{ \tan(t)\sec(t) * (3+\sec(t)) - \tan(t)\sec(t) * \sec(t) }{ (3+\sec(t))^2 }\]

    • 2 years ago
  13. Coolsector Group Title
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    \[f'(t) = \frac{ 3\tan(t)\sec(t) }{ (3+\sec(t))^2 }\]

    • 2 years ago
  14. Coolsector Group Title
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    nothing much more to do ..

    • 2 years ago
  15. v.s Group Title
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    so you expand and simplify?

    • 2 years ago
  16. Coolsector Group Title
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    the final answer ? you can expand the (3+sec(t))^2 but it's not really important ..

    • 2 years ago
  17. Coolsector Group Title
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    i used the quotient rule in order to find the derivative as you can see the procedure

    • 2 years ago
  18. v.s Group Title
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    okaay thank you :)

    • 2 years ago
  19. Coolsector Group Title
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    yw

    • 2 years ago
  20. v.s Group Title
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    i got the aanswer wrong

    • 2 years ago
  21. v.s Group Title
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    i'm doing this online quiz and wen i put in the answer it was wrong

    • 2 years ago
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