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Derivative of (sinx+cosx)secx

  1. Oct 8, 2009 #1
    1. As the title states, I need to fine d/dx(sinx+cosx)secx.

    2. I am given pretty much most of the derivatives and trig functions.

    3. Here's my attempt to solve the question:
    y=(sinx+cos)secx
    y'=(sinx+cosx)(secxtanx)+(secx)(cosx-sinx)
    =sinxsecxtanx+cosxsecxtanx+cosxsecx-sinxsecx
    =secx(sinxtanx+cosxtanx+cosx-sinx)
    and that's where I've gotten up to.


    So now I'm curious if my last step is the final answer, or can I simplify it more. I know some can be switched around but I'm not sure if that would help make the answer. Please advice and thanks for the help in advance!
     
  2. jcsd
  3. Oct 8, 2009 #2

    Office_Shredder

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    Try turning the secants and tangents into sines and cosines before any re-arranging (after taking the derivative though) and see if you get something that you consider 'nicer'. I think most people would accept

    y'=(sinx+cosx)(secxtanx)+(secx)(cosx-sinx)

    As an answer, but obviously I don't know the criteria on which you're graded
     
  4. Oct 8, 2009 #3

    D H

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    That simplifies quite a bit.

    Hint/sanity check: Simplify (sinx+cosx)secx, then differentiate.
     
  5. Oct 8, 2009 #4
    Ok, after reading DH's reply, I tried the question again and here's what I did.
    y=(sinx+cosx)1/cosx
    y'=(sinx+cosx)(-1/sinx)+(1/cosx)(-sinx+cosx)
    =cosx/-sinx + sinx/-sinx + cosx/cosx - sinx/cosx
    =cosx/-sinx - sinx/cosx

    does that seem simple enough for you guys? I wish I were given the answers to check with.
     
  6. Oct 8, 2009 #5

    D H

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    Try simplifying (sinx+cosx)secx again. Hint: What is cos(x)*sec(x)?
     
  7. Oct 8, 2009 #6
    Oh, I see where you're going...
    (sinx)(secx)+1 <----since cosx*secx = cosx*1/cosx
    (sinx)(1/cosx)+1
    [[(cosx)(cosx)-(sinx)(-sinx)]/(cosx)^2]+1
    [(cos2x+sin2x)/cos2x]+1
    (1/cos2x)+1?
    is the +1 supposed to stay?
     
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