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Math Help - Derivative of a inverse trig function... Is my answer wrong?

  1. #1
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    Derivative of a inverse trig function... Is my answer wrong?

    f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)

    Find the derivative then make it into the format:
    <br />
\displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}<br />

    I got it to the form
    <br />
\frac{-(x+7)}{\sqrt{49-x^2}}<br />

    Soooo shouldn't..
    c=7
    m=-1
    n=49
    p=1

    Or have I done something wrong?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Not quite right --note you need the same power top and bottom

    Note 49 -x^2 = (7-x) (7+x)

    you have - [sqrt(x+7)/sqrt((7-x)] = - [(x+7)/(-x+7)]^(1/2)


    p =1/2

    n = 7

    c=7

    m = -1
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  3. #3
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    Wouldn't I then currently have

    -(x+7)/((7-x) (7+x))^(1/2)?

    I think I know where your going but I don't see it :S, Can you explain that?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    x+7 = 7 + x

    so (x+7)/[sqrt(7+x)sqrt(7-x)]=( 7 +x)/[sqrt(7+x)sqrt(7-x)]
    = sqrt(7+x)/sqrt(7-x)
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  5. #5
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    OHh okay, thank you!
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