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Differentiating cotx

  1. Dec 18, 2014 #1
    I keep getting the wrong answer when i try to differentiate cotx..
    this is what i get:
    cotx = 1/tanx =cosx/sinx=cosx ⋅ sin^-1
    so by the product and chain rule we have:
    sinx⋅(sin x)^-1+cos⋅(-1sin^2 x)^-1 ⋅(cosx)^-1

    =

    sinx/sinx - cosx/cosx ⋅ sin^2x
    =1-1/sin^2 x

    where as the correct answer is -1/sin^2x = -csc^2 x

    could someone please tell me where i am going wrong?


    many thanks
    Ryan
     
  2. jcsd
  3. Dec 18, 2014 #2

    Mark44

    Staff: Mentor

    d/dx(cos(x)) = -sin(x). It looks like you have other mistakes as well. For this problem it's probably simpler to use the quotient rule. You don't need to use the chain rule when you do so.
     
  4. Dec 18, 2014 #3

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You have cot(x) = cos(x) * sin-1(x) = u * v

    u = cos (x)
    v = sin-1(x)

    u' = -sin(x)
    v' = -sin-2(x) * cos (x) [from the chain rule]

    d(cot(x))/dx = u * v' + v * u' = -cos2(x)*sin-2(x) - sin(x) * sin-1(x) = -cot2(x) - 1 = -[1 + cot2(x)]

    cot2(x) = cos2(x) / sin2(x)

    1 + cot2(x) = 1 + cos2(x)/sin2(x) = [sin2(x) + cos2(x)] / sin2(x) = 1/sin2(x) = csc2(x)

    -[1 + cot2(x)] = -csc2(x) = d(cot(x))/dx

    Q.E.D.
     
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