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Math Help - Derivatives of Inv Trig Functions

  1. #1
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    Derivatives of Inv Trig Functions

    What am I doing wrong in this problem?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by kid funky fried View Post
    What am I doing wrong in this problem?
    Try this...

    y=\sin^{-1}(x/2)  \iff \sin(y)=\frac{x}{2} by implicit

    \cos(y) \cdot \frac{dy}{dx}=\frac{1}{2} solve for the derivative

    \frac{dy}{dx}=\frac{1}{2 \cos(y)}

    note that \cos(y) =\sqrt{\cos^2(y)}=\sqrt{1-sin^2(y)}=\sqrt{1-(\frac{x}{2})^2}=\frac{1}{2}\sqrt{4-x^2}

    \frac{dy}{dx}=\frac{1}{2 \cos(y)}=\frac{1}{\sqrt{4-x^2}}

    I hope this helps.
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  3. #3
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    y = arcsin^{-1}\left(\frac{x}{2}\right)
    y' = \frac{1}{\sqrt{1 - \left(\frac{x}{2}\right)^{2}}} \cdot \frac{d}{dx} \left(\frac{x}{2}\right)
    y' = \frac{1}{\sqrt{1 - \frac{x^{2}}{4}}} \cdot \frac{1}{2}
    y' = \frac{1}{2 \sqrt{\frac{1}{4}(4 - x^{2})}}
    y' = \frac{1}{2 \cdot \frac{1}{2}\sqrt{4 - x^{2}}} You factor out a 1/2, not a 2 in the denominator
    y' = \frac{1}{\sqrt{4 - x^{2}}}
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  4. #4
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    ha!

    Great, I see now!
    Thanks alot!
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