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Another Deriv

  1. Nov 1, 2006 #1
    (sec4x + 4arctanx^2)=

    (sec4x)(tan4x) + (4)(1/1+x^4)(2x).. Did I derive this correctly??
     
  2. jcsd
  3. Nov 1, 2006 #2

    Office_Shredder

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    Did you use the chain rule on the sec(4x) term?
     
  4. Nov 1, 2006 #3
    you forgot to multiply by [tex] du = 4 [/tex] in the first term
     
  5. Nov 1, 2006 #4
    (sec4x)(4)(tan4x)(4) + (4)(1/1+x^4)(2x)

    or

    16(sec4x)(tan4x) + (4)(1/1+x^4)(2x)??
     
  6. Nov 1, 2006 #5
    or do i keep just 1 4?? like this..

    (sec4x)(tan4x)(4) + (4)(1/1+x^4)(2x)
     
  7. Nov 1, 2006 #6

    Office_Shredder

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    It would only be one four. Let sec(4x)=sec(u).

    Then d(sec4x)/dx = d(secu)/dx = secu*tanu*du/dx

    du/dx = 4, so

    sec(4x)' = 4sec(4x)tan(4x)
     
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