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Trig derivatives

  1. Oct 10, 2005 #1
    im supposed to prove that




    have horisontal tangents for x=0

    i got the derivatives just fine, i proved the first on e with no problem, but what do i do with second? i get to a point where i must prove that Sec(x)Tan(x)=0 but sec x = 1/cos(x) and i can't work with that...i mean how can i divide 1 by something to get zero? it's an asymptote...so what do i do?

    i know how to get derivatives, i don't know how to prove that the second one is having the horisontal tangent.
  2. jcsd
  3. Oct 11, 2005 #2


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    y = secx
    y' = secx.tanx = sinx/cos²x

    at x = 0, sinx = 0, cosx = 1

    y' = 0/1² = 0
  4. Oct 11, 2005 #3
    okay...i thought that when you have secx*tanx=0 you have to set them like when you have parenthesis...each one at a time. That was probably a better explanation than most teachers would give. Thank you.

    What i was trying to do is prove that either tan or sec is some value and the second one is zero, so zero*value=0 but it doesn't work that way.

    can i ask though...why was i wrong? i mean if you do (x)(X+1)=0 you're defenetly not wrong to set either one equl to zero...why wasn't my way working? (To be honest im more interested on what went wrong than what the answer is...i can always copy that from the book but i don't want to).
  5. Oct 11, 2005 #4


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    As far as I can see, you weren't wrong!

    You had,

    y' = secx*tanx

    at x = 0, secx= 1/cosx = 1/1 = 1, and tanx = 0,


    y' = 1*0 = 0

    Waht you have is secx = 1 and the other one, tanx = 0
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