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Derivatives of inverse trig functions

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    find an equation of the tangent line to graph of f at the indicated point.

    f(x) = arcsin2x
    (u'/sqrt 1-u^2)(2)


    2. The attempt at a solution
    (1/sqrt 1-2x^2)(2)
    I got the answer from calcchat but I don't understand where the 1 and 2 came from?
     
  2. jcsd
  3. Oct 2, 2007 #2

    rock.freak667

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    Let [tex]y=sin^{-1}2x=> siny=2x[/tex] can you do implicity Differentiation?

    or here is a formula to know

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

    and because you know that formula...you can use the chain rule
     
  4. Oct 2, 2007 #3
    Okay I thought the formula was....
    u'/sqrt 1-u^2
    so can u bring it up so its 1-(1-x)^-1/2?
     
  5. Oct 2, 2007 #4

    rock.freak667

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

    because [tex]\frac{a}{b^n} = a*b^{-n}[/tex]
     
  6. Oct 3, 2007 #5

    HallsofIvy

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    Obviously, if u= x (and the differentiation is with respect to x) then u'= 1 so the two formulas are the same. The formula you give is more general. If u is any function of x (and the differentiation is with respect to x) then
    [tex]\frac{d arcsin(u(x))}{dx}= \frac{ \frac{du}{dx}}{\sqrt{1- u^2(x)}}[/tex]
     
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