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Math Help - Square root derivative

  1. #1
    Super Member Bacterius's Avatar
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    Square root derivative

    Hi, I'm trying to derivate this function : f(x) = \sqrt{2x - 1}. I'm applying this principle : if f = a^n then f' = na^{n - 1}. So there I go :

    f(x) = \sqrt{2x - 1}

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

    Now I derivate :

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

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

    f'(x) = \frac{1}{2} \frac{1}{\sqrt{2x - 1}}

    f'(x) = \frac{1}{2 \sqrt{2x - 1}}

    But the book says that's not right and the answer should be :

    f'(x) = \frac{2}{2 \sqrt{2x - 1}} = \frac{1}{\sqrt{2x - 1}}

    Does anyone see when I went wrong ? Thanks a lot ...
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Bacterius View Post
    Hi, I'm trying to derivate this function : f(x) = \sqrt{2x - 1}. I'm applying this principle : if f = a^n then f' = na^{n - 1}. So there I go :

    f(x) = \sqrt{2x - 1}

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

    Now I derivate :

    f'(x) = \frac{1}{2} (2x - 1)^{\frac{-1}{2}} <<<<<<< here! You forgot to apply the chain rule

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

    f'(x) = \frac{1}{2} \frac{1}{\sqrt{2x - 1}}

    f'(x) = \frac{1}{2 \sqrt{2x - 1}}

    But the book says that's not right and the answer should be :

    f'(x) = \frac{2}{2 \sqrt{2x - 1}} = \frac{1}{\sqrt{2x - 1}}

    Does anyone see when I went wrong ? Thanks a lot ...
    f(x) = (2x - 1)^{\frac{1}{2}}

    f'(x) = \underbrace{\frac{1}{2} (2x - 1)^{\frac{-1}{2}}}_{\text{der. of the square-root}} \cdot \underbrace{2}_{\text{der. of the radicand}}
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  3. #3
    Super Member Bacterius's Avatar
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    Thanks a lot !
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