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Homework Help: Derivative of inverse trig function

  1. Jun 3, 2006 #1
    Hi, can anyone check my work?

    [tex]y=tan^-^1(x-(1+x^2)^\frac{1}{2})[/tex]

    [tex]y^1=\frac{1}{1+u^2}\frac{dy}{du}[/tex]

    let u = (x-(1+x^2))

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

    thanks
     
    Last edited: Jun 3, 2006
  2. jcsd
  3. Jun 3, 2006 #2

    arildno

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    Dearly Missed

    Your post is littered with errors. :mad:
    Edit: I've edited in for your forgotten square root sign. :mad::mad:
    Here's how you may do it:
    Set [itex]Y(u)=\tan^{-1}(u),U(x)=(x-\sqrt{1+x^{2}}), y(x)=Y(U(x))[/tex]
    Thus, we have:
    [tex]\frac{dy}{dx}=\frac{dY}{du}\mid_{u=U(x)}\frac{dU}{dx}[/tex]
    Or:
    [tex]\frac{dY}{du}\mid_{u=U(x)}=\frac{1}{1+U(x)^{2}},\frac{dU}{dx}=1-\frac{x}{\sqrt{1+x^{2}}}[/tex]
    Or, finally:
    [tex]\frac{dy}{dx}=(1-\frac{x}{\sqrt{1+x^{2}}})\frac{1}{1+(x-\sqrt{1+x^{2}})^{2}}[/tex]
     
    Last edited: Jun 3, 2006
  4. Jun 3, 2006 #3
    oops..i made an error in my first equation..i edited it..can you check it now?
     
  5. Jun 4, 2006 #4

    HallsofIvy

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    If you edited this after arildno looked at it, you are still missing a square root:
    [tex]y=tan^-^1(x-(1+x^2)^\frac{1}{2})= tan^{-1}(u)[/tex]
    for u= (x- (1+ x2)1/2).
     
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