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Math Help - Derivative

  1. #1
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    Derivative

    How do I get the correct derivative for this function?

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

    I got the first part of \frac{2}{\sqrt{x+1}} but cant seemed to get the second part of it using the quotient rule \frac{2x-1}{2(x+1)^{3/2}}

    y' in the answer's is supposedly

    y ' = \frac{2}{\sqrt{x+1}}-\frac{2x-1}{2(x+1)^{3/2}}

    Can someone please show me how to get to that using the quotient rule?
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  2. #2
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    Quote Originally Posted by dwat View Post
    How do I get the correct derivative for this function?

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

    I got the first part of \frac{2}{\sqrt{x+1}} but cant seemed to get the second part of it using the quotient rule \frac{2x-1}{2(x+1)^{3/2}}

    y' in the answer's is supposedly

    y ' = \frac{2}{\sqrt{x+1}}-\frac{2x-1}{2(x+1)^{3/2}}

    Can someone please show me how to get to that using the quotient rule?
    y = \frac{2x - 1}{\sqrt{x + 1}} = \frac{2x - 1}{(x + 1)^{\frac{1}{2}}}.


    Using the Quotient rule:

    \frac{dy}{dx} = \frac{(x + 1)^{\frac{1}{2}}\frac{d}{dx}(2x - 1) - (2x - 1)\frac{d}{dx}\left((x + 1)^{\frac{1}{2}}\right)}{\left((x + 1)^{\frac{1}{2}}\right)^2}

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


    You could try to simplify that or write it with radicals if you wish...
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