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Math Help - Derivitive of a Sqrt of a rational function

  1. #1
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    Derivitive of a Sqrt of a rational function

    f(x) = sqrt [(2x+1) / (2x-1)]

    find f'(x)

    I'm using chain rule, but I'm not sure to bring the ^(1/2) to both top and bottom terms and then apply the quotiant rule.

    any help wouldbe wonderful.
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  2. #2
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    Quote Originally Posted by youmuggles View Post
    f(x) = sqrt [(2x+1) / (2x-1)]

    find f'(x)

    I'm using chain rule, but I'm not sure to bring the ^(1/2) to both top and bottom terms and then apply the quotiant rule.

    any help wouldbe wonderful.
    Yes,
    take the square root of the numerator and denominator,
    use power of 0.5 for both,
    apply the quotient rule,
    use the chain rule for the derivatives of numerator and denominator.
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  3. #3
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    Quote Originally Posted by youmuggles View Post
    f(x) = sqrt [(2x+1) / (2x-1)]

    find f'(x)

    I'm using chain rule, but I'm not sure to bring the ^(1/2) to both top and bottom terms and then apply the quotiant rule.

    any help wouldbe wonderful.
    I find that the chain rule makes more sense when you use a u-substitution. Just set the 'inside' function equal to u:

    u=\frac{2x+1}{2x-1}

    y=\sqrt{u}

    Since u is a function of x, the chain rule gives:

    \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

    \frac{dy}{du}=\frac{1}{2\sqrt{u}}

    \frac{du}{dx}=\frac{d}{dx}\frac{2x+1}{2x-1}

    Do you see how this works?
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  4. #4
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    If you want to consider other simplifications,
    you can try a few tricks such as..


    \sqrt{\frac{2x+1}{2x-1}}=\frac{\sqrt{2x+1}}{\sqrt{2x-1}}

    =\frac{\sqrt{2x+1}\sqrt{2x-1}}{\sqrt{2x-1}\sqrt{2x-1}}

    =\frac{\sqrt{(2x+1)(2x-1)}}{2x-1}=\frac{\sqrt{4x^2-1}}{2x-1}

    Then differentiate that.

    Or...

    \frac{\sqrt{2x+1}}{\sqrt{2x-1}}=(2x+1)^{\frac{1}{2}}(2x-1)^{-\frac{1}{2}}

    Let y=2x-1

    \frac{dy}{dx}=2

    \frac{d}{dx}[(y+2)^{\frac{1}{2}}y^{-\frac{1}{2}}]

    =\frac{dy}{dx}\frac{d}{dy}[(y+2)^{\frac{1}{2}}y^{-\frac{1}{2}}]=2\frac{d}{dy}[(y+2)^{\frac{1}{2}}y^{-\frac{1}{2}}]

    Then use the "product rule".
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  5. #5
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    [Math]\left(\sqrt{\frac{2x+1}{2x-1}}\right)'=\left(\frac{\sqrt{2x+1}}{\sqrt{2x-1}}\right)'
    [/tex] = \frac{\left(\sqrt{2x+1}\right)'\left(\sqrt{2x-1}\right)-\left(\sqrt{2x+1}\right)\left(\sqrt{2x-1}\right)'}{\left(\sqrt{2x-1}\right)^2} = \dfrac{\frac{\sqrt{2x-1}}{\sqrt{2x+1}}-\frac{\sqrt{2x+1}}{\sqrt{2x-1}}}{2x-1} = \frac{\sqrt{2x-1}\sqrt{2x-1}-\sqrt{2x+1}{\sqrt{2x+1}}}{(2x-1)\sqrt{2x+1}\sqrt{2x-1}} = \dfrac{(2x-1)-(2x+1)}{\sqrt{(2x-1)^3}\sqrt{2x+1}}  = \boxed{\dfrac{-2}{\sqrt{(2x-1)^3}\sqrt{2x+1}}}.
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