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Proof of d/dx arcsin(x)

  1. Sep 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that:
    [tex]\frac{d}{dx}arcsin(x) = \frac{1}{\sqrt{1-x^{2}}}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex]y=arcsin(x)[/tex]
    [tex]sin(y)=sin(arcsin(x))[/tex]
    [tex]sin(y)=x[/tex]
    [tex]\frac{d}{dx}(sin(y)=x)[/tex]
    [tex]cos(y)\frac{dy}{dx}=1[/tex]
    [tex]\frac{dy}{dx}=\frac{1}{cos(y)}[/tex]
    Using a triangle:
    [tex]cos(y)=\sqrt{1-x^2}[/tex]
    [tex]\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}[/tex]
    [tex]\frac{d}{dx}arcsin(x) = \frac{1}{\sqrt{1-x^{2}}}[/tex]

    We were shown these in class before starting trig-sub and all that, and the professor said that there are proofs by differentiation. I don't know how to explain the steps where I determine that the cosine of the angle y is equal to sqrt(1-x^2) without drawing out a triangle. Is there some way? Other than that I think I have it figured out and I can do the same for the other functions.
     
  2. jcsd
  3. Sep 9, 2011 #2
    Well, certainly,

    [tex]x=sin(y)[/tex]

    In addition,

    [tex]sin^2y+cos^2y=1[/tex]

    So it certainly follows that,

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

    You can see that it follows naturally (by substitution) that,

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

    And we're done!
     
  4. Sep 9, 2011 #3
    Oh haha. That really should have occurred to me considering that I used the pyth. theorem to find cos(y).

    Thanks
     
  5. Sep 9, 2011 #4

    HallsofIvy

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    Since you refer to "Using a triangle", you can also do it this (equivalent) way:
    imagine a right triangle triangle having "opposite side" of length x and "hypotenuse" of length 1, so that sin(y)= x/1= x. Then, by the Pythgorean theorem, the "near side" has length [itex]\sqrt{1- x^2}[/itex]. From that, [itex]cos(y)= \sqrt{1- x^2}[/itex].
     
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