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Math Help - Need help finding the derivative.

  1. #1
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    I need to find the derivative of

     x-2 \sqrt{x-1}

    I knew to use the power rule, but I was coming up with
     1/2(x-1)^{-1/2} Is this the correct answer?

    Thanks
    Jason
    Last edited by mr fantastic; June 22nd 2009 at 02:03 AM. Reason: Merged posts, fixed latex
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    I knew to use the power rule, but I was coming up with
     1/2(x-1)^-1/2 Is this the correct answer?
    The derivative (by power rule & chain rule) is 1-\left(x-1\right)^{-1/2}=1-\frac{1}{\sqrt{x-1}}, where the derivative of \sqrt{x-1} is \frac{1}{2}\left(x-1\right)^{-1/2}=\frac{1}{2\sqrt{x-1}}
    Last edited by mr fantastic; June 22nd 2009 at 02:04 AM. Reason: Deleted some bits due to original ambiguity in OP
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    The derivative (by power rule & chain rule) is 1-\left(x-1\right)^{-1/2}=1-\frac{1}{\sqrt{x-1}}, where the derivative of \sqrt{x-1} is \frac{1}{2}\left(x-1\right)^{-1/2}=\frac{1}{2\sqrt{x-1}}
    How did you know to use the chain rule?
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  4. #4
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    chain rule is present at any moment.

    for example if you want to differentiate a simple function as x^2, then that is 2x\cdot(x)'=2x, but that it's quite simple and then irrelevant to write, but having \sqrt{x-1} we can think this as the composition by letting f(x)=\sqrt x and g(x)=x-1, so by differentiating f\big(g(x)\big) we get f'\big(g(x)\big)\cdot g'(x). According to this we have that \sqrt{x-1} can be expressed as h(x)=f\big(g(x)\big)=\sqrt{x-1}, and then by differentiating we get (following up on what I said before) h'(x)=\frac1{2\sqrt{x-1}}\cdot(x-1)'=\frac1{2\sqrt{x-1}}.
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