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Have I done this right?

  1. Jun 22, 2005 #1
    To find dy/dx for

    [tex] y = ln(sin^{-1}(x)) [/tex]

    I did this:

    [tex]
    y = ln(sin^{-1}(x))
    [/tex]

    so

    [tex]
    e^y = (sin^{-1}(x))
    [/tex]

    and

    [tex]
    e^y \frac {dy}{dx} = \frac {1} {\sqrt{1-x^2}}
    [/tex]

    then

    [tex]
    \frac {dy}{dx} = \frac {1}{\sqrt {1-x^2} * e^y}
    [/tex]

    [tex]
    = \frac {1}{\sqrt {1-x^2} * e^{ln(sin^{-1}(x))}}
    [/tex]

    [tex]
    = \frac {1}{\sqrt {1-x^2} * (sin^{-1}(x))}}
    [/tex]

    I think that its all valid implicit differentiation but I m not 100% confident about it, please help.

    :frown:
     
    Last edited: Jun 22, 2005
  2. jcsd
  3. Jun 22, 2005 #2
    It looks fine to me, and the result checks out with ordinary differentiation. Which part were you uncertain about ?
     
  4. Jun 22, 2005 #3

    Zurtex

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    Homework Helper

    You could have just as easily applied the chain rule but that seems to be fine :smile:
     
  5. Jun 22, 2005 #4
    Just the validity of transferring y to differentiate both sides, I wasn't sure that I was differentiating the same equation, probably over thinking it.

    Thanks heaps though because I think I am starting to become more capable with it but I don't want to be overconfident in mistakes.

    :cool:
     
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