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2 answers?

  1. Sep 17, 2011 #1
    d/dx (sin-1x) = [itex]\frac{1}{\sqrt{1-x2}}[/itex]

    But using chain rule, I got:

    dy/dx = -sin-2xcosx

    Why???
     
  2. jcsd
  3. Sep 17, 2011 #2

    gb7nash

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    sin-1x is not equal to [itex]\frac{1}{\sin x}[/itex]. sin-1x stands for the arcsin of x, or more specifically, the angle y that satisfies sin y = x.

    I've always personally disliked the notation sin-1x, but it is what it is.
     
  4. Sep 17, 2011 #3

    HallsofIvy

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    If [itex]y= sin^{-1}(x)[/itex] then [itex]x= sin(y)[/itex].

    [tex]\frac{dx}{dy}= cos(y)[/tex]

    so that
    [tex]\frac{dy}{dx}= \frac{1}{cos(y)}[/tex]

    but [tex]y= sin^{-1}(x)[/tex] so [tex]cos(sin^{-1}(x))= \sqrt{1- sin^2(sin^{-1}(x))}= \sqrt{1- x^2}[/tex]

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

    As gb7nash said, the "-1", applied to functions denotes the "inverse function" (the inverse for function composition), not the reciprocal (the inverse for multiplication).
     
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