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Calculus I - Derivitive

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    sin2x / (sin2x + secx) d/dx = ?


    2. Relevant equations

    d/dx (f/g) = (g d/dx f - f d/dx g) / g^2
    d/dx sec u = sec u tan u d/dx u
    d/dx cos u = -sin u d/dx u
    d/dx sin u = cos u d/dx u

    3. The attempt at a solution

    d/dx y= [(sin2x + secx) d/dx (sin2x) - sin2x d/dx (sin2x + secx)] / (sin2x + secx)^2

    = [(sin2x + secx) cosx d/dx (2x) - sin2x d/dx (sin2x) + d/dx secx)] / (sin2x + secx)^2

    = [(sin2x + secx) cosx (2) - sin2x cos2x (2) + (sec2x*tan2x) 2] / (sin2x + secx)^2

    = [2(sin2x + secx) cosx - 2sin2x cos2x + 2(sec2x*tan2x)] / (sin2x + secx)^2

    ^ That is my final answer, can anyone confirm for me?
     
  2. jcsd
  3. Oct 26, 2009 #2

    Mark44

    Staff: Mentor

    Mistakes in next line. d/dx(sin2x) = cos2x d/dx(2x) = 2 cos2x.
    d/dx(secx) = secx * tanx
     
    Last edited: Oct 26, 2009
  4. Oct 26, 2009 #3
    On our given formulae sheet sec x is equal to sec x * tan x :S
     
  5. Oct 26, 2009 #4

    Mark44

    Staff: Mentor

    OK, my mistake, which I have fixed, but your formula sheet should say d/dx(secx) = secx * tan x, not sec x = secx * tanx.
     
  6. Oct 26, 2009 #5
    Yes sir, that is correct. But I am skipping a step.

    d/dx (sin2x + secx)

    is equal to d/dx sin2x + d/dx secx

    I just failed to put that step in, would you agree with this?
     
  7. Oct 26, 2009 #6

    Mark44

    Staff: Mentor

    Yes, I agree with that, but what do you get for d/dx sin2x + d/dx secx?

    Also, in one of your lines you had a numerator of
    [(sin2x + secx) cosx d/dx (2x) - sin2x d/dx (sin2x) + d/dx secx)]
    You're missing a left parenthesis somewhere; it should be right between -sin2x and d/dx. IOW, like so: [(sin2x + secx) cosx d/dx (2x) - sin2x (d/dx (sin2x) + d/dx secx)]
    ...............................^
    ...............................|
     
  8. Oct 26, 2009 #7
    Thank you.
    d/dx sin2x + d/dx secx = cos2x (d/dx 2x) + tanx*secx

    Ahh, I see my fault.. the correct solution would be:



    = [2(sin2x + secx) cosx - 2sin2x cos2x + (secx*tanx)] / (sin2x + secx)^2

    Thank you, I always make stupid mistakes. Is this the answer you are thinking on?
     
  9. Oct 26, 2009 #8

    Mark44

    Staff: Mentor

    No, parts are wrong. Here's what I get:
    d/dx[sin2x/(sin2x + sec x)] = [(sin2x + secx)(2cos2x) - (sin2x)(2cos2x + secxtanx]/(sin2x + secx)^2
    = [2sin2x cos2x + 2secx cos2x - 2sin2x cos2x - sin2x secx tanx]/(sin2x + secx)^2
     
  10. Oct 26, 2009 #9

    Mark44

    Staff: Mentor

    A piece of positive criticism about your work is that you are setting things up correctly when you start in on the quotient rule. This makes it very easy to follow what you're doing. A little more practice and you'll be able to do this with much fewer mistakes.
     
  11. Oct 27, 2009 #10
    Ahh yes I understand, just working it out on paper for myself now. Thank you
     
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