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rainbow22

  • 3 years ago

find y" if y= x tan x

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  1. bahrom7893
    • 3 years ago
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    y' = x*(Secx)^2+tanx

  2. rainbow22
    • 3 years ago
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    I found that y' is [x sec^2 x + tan x]

  3. bahrom7893
    • 3 years ago
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    bb in 10 mins

  4. rainbow22
    • 3 years ago
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    I don't know how to do the next step...

  5. bahrom7893
    • 3 years ago
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    y'' = x*2*Secx*(SecxTanx) + (Secx)^2 + (Secx)^2

  6. bahrom7893
    • 3 years ago
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    simplify that

  7. rainbow22
    • 3 years ago
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    how did you arrive at that answer?

  8. ilikephysics2
    • 3 years ago
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    you need the 2nd derivative

  9. bahrom7893
    • 3 years ago
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    use product and power rules on x*(Secx)^2

  10. bahrom7893
    • 3 years ago
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    derivative of (Secx)^2 is 2*Secx*(SecxTanx)

  11. rainbow22
    • 3 years ago
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    why wouldn't it be 2x (sec x tan x)

  12. rainbow22
    • 3 years ago
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    Is it because you have to use something called a chain rule?

  13. bahrom7893
    • 3 years ago
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    Sorry was away. Yes it is the chain rule and the product rule: derivative of u^a with respect to x is: a * (u)^(a-1) * (du/dx)

  14. rainbow22
    • 3 years ago
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    Can you not do it with the chain rule it's long and tedious but..

  15. bahrom7893
    • 3 years ago
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    u can't do it with any other rule

  16. bahrom7893
    • 3 years ago
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    it's (Secx)^2 you have to use the chain rule.

  17. rainbow22
    • 3 years ago
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    Oh, there was a way to use the constant multiple rule the sum/difference rule and the product rule to get the answer :( That's the way my book shows it. Because I'm learning the chain rule in the nxt ch.

  18. bahrom7893
    • 3 years ago
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    x IS NOT a constant......................

  19. bahrom7893
    • 3 years ago
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    neither is (Secx)^2

  20. rainbow22
    • 3 years ago
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    pg 91 number 27

  21. bahrom7893
    • 3 years ago
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    ohhhhhhhhhhhhh

  22. rainbow22
    • 3 years ago
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    sorry, should have put that up..

  23. Freyes
    • 3 years ago
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    *waits for answer*

  24. bahrom7893
    • 3 years ago
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    Man I'd kill them for making me use the product rule on that.. (Secx)^2 = Secx*Secx = Secx*(SecxTanx) + (SecxTanx)*Secx

  25. rainbow22
    • 3 years ago
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    now... uhhh...

  26. bahrom7893
    • 3 years ago
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    Ok... let's start over.. One last time... http://www.youtube.com/watch?v=0Ox3V-1aI38

  27. bahrom7893
    • 3 years ago
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    ^joke.. anyway.. y' = x*(Secx)^2+tanx y'' = x*[(Secx)^2] ' + 1*(Secx)^2 + (Secx)^2

  28. rainbow22
    • 3 years ago
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    I got lost

  29. bahrom7893
    • 3 years ago
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    Where?

  30. rainbow22
    • 3 years ago
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    why is there two sec x ^2

  31. bahrom7893
    • 3 years ago
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    There are 3 Secx^2.. which 2 are u talking about?

  32. rainbow22
    • 3 years ago
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    how did tan x become 2 (sec x ^2)

  33. bahrom7893
    • 3 years ago
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    Derivative of tangent is (Secx)^2.. you should memorize that. And derivative of cotangent is -(Cscx)^2

  34. rainbow22
    • 3 years ago
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    yeah.. but you have 2 of sec ^2

  35. bahrom7893
    • 3 years ago
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    Okay... I was really hoping that we wouldn't start over again, but lets start over.. y' = x*(Secx)^2+tanx y '' then would be: {x*(Secx)^2}' + {tanx}' are you following so far?

  36. rainbow22
    • 3 years ago
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    yes

  37. bahrom7893
    • 3 years ago
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    Now look at the first term: {x*(Secx)^2}' = *using the product rule* = x*{(Secx)^2}' + {(Secx)^2}*1 right?

  38. rainbow22
    • 3 years ago
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    yes so sec (sec)' sec (sec)'

  39. rainbow22
    • 3 years ago
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    and sec' is euqal to sec*tan

  40. bahrom7893
    • 3 years ago
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    Not yet.. don't rush

  41. rainbow22
    • 3 years ago
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    okay

  42. bahrom7893
    • 3 years ago
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    Now we need {(Secx)^2}' {(Secx)^2}' = {Secx*Secx}' = *using product rule* = Secx *(Secx)' + (Secx)'*Secx = = Secx * (SecxTanx) + (SecxTanx) * Secx

  43. bahrom7893
    • 3 years ago
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    Again.. how do we know that (Secx)' = SecxTanx.. you memorize it. (Cscx)' = -CscxCotx btw

  44. rainbow22
    • 3 years ago
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    gots that

  45. bahrom7893
    • 3 years ago
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    So {(Secx)^2}' = 2(Secx)^2(Tanx) Plug that in here: {x*(Secx)^2}' = *using the product rule* = x*{(Secx)^2}' + {(Secx)^2}*1 = = x * 2(Secx)^2(Tanx) + (Secx)^2

  46. bahrom7893
    • 3 years ago
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    Are you still following?

  47. rainbow22
    • 3 years ago
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    er... hang on for a sec

  48. bahrom7893
    • 3 years ago
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    ok

  49. rainbow22
    • 3 years ago
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    wouldn't it be... x sec sec (sec tan)(sec tan) ??

  50. bahrom7893
    • 3 years ago
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    No.. Ok let's try Latex maybe then it's going to be easier to see.

  51. bahrom7893
    • 3 years ago
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    Did you follow how we got: \[{(Secx)^2}' = 2(Secx)^2(Tanx)\]

  52. bahrom7893
    • 3 years ago
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    Darn it the rest didn't show up

  53. rainbow22
    • 3 years ago
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    sec^2 = sec sec sec (sec tan) + sec (sec tan) sec^2 (sec tan) + sec^2 (sec tan) 2 sec^2 + 2 sec tan?

  54. bahrom7893
    • 3 years ago
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    sec^2 = sec sec sec (sec tan) + sec (sec tan) <---Here you're just multiplying it out. sec^2 (tan) + sec^2 (tan)

  55. bahrom7893
    • 3 years ago
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    sec^2 (tan) + sec^2 (tan) = 2(sec^2 (tan))... something plus something is 2 of that something

  56. bahrom7893
    • 3 years ago
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    Did you follow?

  57. rainbow22
    • 3 years ago
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    yes

  58. bahrom7893
    • 3 years ago
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    Okay.. THUS: \[(Sec^2x)'=2(Sec^2x)(Tanx)\]

  59. rainbow22
    • 3 years ago
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    yep and tan x -> sec^2

  60. bahrom7893
    • 3 years ago
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    Now:.. uhmm i forgot what the question was.. lol

  61. rainbow22
    • 3 years ago
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    and.. on the constant multiple rule.. just 'add in' the extra x in the beginning?

  62. bahrom7893
    • 3 years ago
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    do not forget that x is not a constant, x is a variable, so you have to use the product rule

  63. rainbow22
    • 3 years ago
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    uh.....?

  64. bahrom7893
    • 3 years ago
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    \[2, \pi, e, 1/2, -0.005\] ^ Those are constants. x is a variable.

  65. rainbow22
    • 3 years ago
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    soo how would you .. take out the x?

  66. bahrom7893
    • 3 years ago
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    Not exactly. \[y' = x*(Secx)^2+tanx\] Look at the first term: \[x*(Secx)^2\]that's just a product rule: \[f(x)*g(x)\] where \[f(x)=x\] and \[g(x)=(Secx)^2\]

  67. rainbow22
    • 3 years ago
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    then.....

  68. bahrom7893
    • 3 years ago
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    So \[(f(x)g(x))'=f(x)*g'(x)+f'(x)*g(x)\] or \[(x*(Secx)^2)'=x*(Sec^2x)'+x'*(Sec^2x)=x*2(Sec^2x)(Tanx)+1*(Sec^2x)\]

  69. bahrom7893
    • 3 years ago
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    Because earlier we showed that: \[{(Secx)^2}' = 2(Secx)^2(Tanx)\]

  70. rainbow22
    • 3 years ago
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    oh... that was the last step right? (cuz I THINK I got the answer)

  71. bahrom7893
    • 3 years ago
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    No that's not the last step.. That's only the first term

  72. rainbow22
    • 3 years ago
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    -.- but then tan'-> sec^2

  73. rainbow22
    • 3 years ago
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    WAIT wouldn't it be 0+sec x ^2

  74. rainbow22
    • 3 years ago
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    wait.. I got it nvrmind confused it with constant

  75. bahrom7893
    • 3 years ago
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    Yes \[y' = x*(Secx)^2+tanx\] The derivative of the first term was: \[(x*Sec^2x)'=x*2(Sec^2x)(Tanx)+1*(Sec^2x)\] The derivative of the second term is: \[(Tanx)'=Sec^2x\]

  76. bahrom7893
    • 3 years ago
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    So the whole thing is: \[y''=2x(Sec^2x)(Tanx)+(Sec^2x)+(Sec^2x)\]

  77. rainbow22
    • 3 years ago
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    I got it! :D

  78. rainbow22
    • 3 years ago
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    and then you can put sec^2 sec^2 as 2(sec^2)

  79. bahrom7893
    • 3 years ago
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    YES!

  80. rainbow22
    • 3 years ago
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    and that's the answer :) THANK YOU SO MUCH!!!!!!!!!!!!!!!!!!!!!

  81. rainbow22
    • 3 years ago
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    lol :) thank you!

  82. bahrom7893
    • 3 years ago
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    np lol.. thanks for sticking to the question

  83. rainbow22
    • 3 years ago
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    I'm really in awe of your patience.. and of course, your skills

  84. rainbow22
    • 3 years ago
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    thanks again!

  85. bahrom7893
    • 3 years ago
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    lol you're welcome.. just practice more

  86. rainbow22
    • 3 years ago
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    will do

  87. rainbow22
    • 3 years ago
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    adios.

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