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Thread: y=(x^2+2x-1)^(1/2) find y'

  1. #1
    Super Member bigwave's Avatar
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    y=(x^2+2x-1)^(1/2) find y'

    $\displaystyle y=\sqrt{x^2+2x-1}$ find $\displaystyle y^,$ answer in book is $\displaystyle \frac{x+1}{y} $ but don't see how?

    I did this...

    $\displaystyle y^, = \frac{1}{2}(x^2+2x-1)^{-\frac{1}{2}}(2x+2)

    \Rightarrow
    \frac
    {x+1}
    {(x^{2}+2x-1)^{\frac{1}{2}}}
    $
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  2. #2
    Super Member bigwave's Avatar
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    never mind I see what is in the denominator is y
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