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reeny1996

  • 2 years ago

if F(x)=tanx, then fprime(pi/6)=

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  1. Shadowys
    • 2 years ago
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    Have you differentiated F(x) and then sub'ed \(\frac{\pi}{6}\) into it?

  2. waterineyes
    • 2 years ago
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    Do you know the derivative of tan(x) ??

  3. reeny1996
    • 2 years ago
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    i kno the differvative is secxtanx but i dont know where to go frm there :/

  4. waterineyes
    • 2 years ago
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    Yes, than just replace x by pi/6..

  5. reeny1996
    • 2 years ago
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    (2\[\sqrt{3}\]/3 and \[\sqrt{3}\]/3

  6. waterineyes
    • 2 years ago
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    \[F'(\frac{\pi}{6}) = \sec(\frac{\pi}{6}) \cdot \tan(\frac{\pi}{6})\]

  7. reeny1996
    • 2 years ago
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    igot tht part

  8. waterineyes
    • 2 years ago
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    What is the value of : \[\sec (\frac{\pi}{6})\]

  9. reeny1996
    • 2 years ago
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    2root3/3

  10. waterineyes
    • 2 years ago
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    Yes and what is for tan(pi/6) = ??

  11. reeny1996
    • 2 years ago
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    root3/3

  12. waterineyes
    • 2 years ago
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    Wait, derivative of tan(x) isn't it sec^2(x) ??

  13. reeny1996
    • 2 years ago
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    its not ?

  14. waterineyes
    • 2 years ago
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    \[\frac{d}{dx} \tan(x) = \sec^2(x)\]

  15. reeny1996
    • 2 years ago
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    thts y im getting the wrong answer

  16. reeny1996
    • 2 years ago
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    thank you

  17. waterineyes
    • 2 years ago
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    You were doing the derivative of sec(x)..

  18. reeny1996
    • 2 years ago
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    yea

  19. waterineyes
    • 2 years ago
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    \[\frac{d}{dx}\sec(x) = \sec(x) \cdot \tan(x)\] It is okay and sorry from my side too, I also got confused.. And you are welcome dear..

  20. reeny1996
    • 2 years ago
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    ok idk how to combine the 2 radicals

  21. waterineyes
    • 2 years ago
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    Where ??

  22. reeny1996
    • 2 years ago
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    sec^2

  23. waterineyes
    • 2 years ago
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    \[F'(\frac{\pi}{6}) = \sec^2(\frac{\pi}{6})\]

  24. reeny1996
    • 2 years ago
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    yea

  25. reeny1996
    • 2 years ago
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    |dw:1352970940060:dw|

  26. waterineyes
    • 2 years ago
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    \[\implies \frac{2 \sqrt{3}}{3} \times \frac{2\sqrt{3}}{3} \implies ??\]

  27. waterineyes
    • 2 years ago
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    This is simple multiplication and division now...

  28. reeny1996
    • 2 years ago
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    i know but im not getting the right answer

  29. reeny1996
    • 2 years ago
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    4

  30. waterineyes
    • 2 years ago
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    What are you getting tell me first...

  31. waterineyes
    • 2 years ago
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    Its not 4... You are missing something.

  32. waterineyes
    • 2 years ago
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    \[\sqrt{3} \times \sqrt{3} = ??\]

  33. reeny1996
    • 2 years ago
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    2 times 2=4 root3 times root 3= 3 4 times 3 =12 12 divide3 =4

  34. waterineyes
    • 2 years ago
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    12 divided by 3 ??? Are you sure in denominator it is 3 ??

  35. reeny1996
    • 2 years ago
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    im not too good with my unit circle

  36. waterineyes
    • 2 years ago
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    Dear, what will you get when you will square 3 ??

  37. reeny1996
    • 2 years ago
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    9

  38. waterineyes
    • 2 years ago
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    Check the colored things here : \[\frac{2 \sqrt{3}}{\color{red}{3}} \times \frac{2 \sqrt{3}}{\color{red}{3}} = ??\]

  39. reeny1996
    • 2 years ago
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    9

  40. waterineyes
    • 2 years ago
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    Yes it is 12 divided by 9 and not 3.. So what will you get now ??

  41. reeny1996
    • 2 years ago
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    ohhhhhh

  42. reeny1996
    • 2 years ago
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    4/3

  43. reeny1996
    • 2 years ago
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    i was thinking addition

  44. reeny1996
    • 2 years ago
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    my mistake...srry if i took up yhur time

  45. waterineyes
    • 2 years ago
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    It is okay but be careful from next time... These are called "Silly Mistakes..." Ok, now concentrate more on new problems...

  46. waterineyes
    • 2 years ago
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    It is okay..

  47. reeny1996
    • 2 years ago
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    thanks a lot :)))

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