mmu

find dy/dx of y=1/2 cot^7(-x^2)

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asked by afiq
  1. Which of these is your school SUBJECT?

    http://www.google.com/#q=mmu

  2. use the chain rule. let

    v = -x^2
    u = cot v
    y = 1/2 u^7

    y' = 7/2 u^6 u'
    u' = -csc^2 v v'
    v' = -2x

    So,

    dy/dx = 7/2 cot^6(-x^2) * -csc^2(-x^2) * -2x
    = 7x cot^2(-x^2) csc^2(-x^2)
    or, since
    cot(-x) = -cotx
    csc(-x) = -csc(x)

    7x cot^2(x^2) csc^2(x^2)

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    posted by Steve

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