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Math Help - Derivative of f(x) = x + sqrt(x)

  1. #1
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    Derivative of f(x) = x + sqrt(x)

    Find f'(x) for the function f(x) = x + \sqrt{x}

    This is Calc 1, so we have to do this the long way for a little while longer, so I have to use the formula

    f'(x) = \displaystyle \lim_{h\to 0 } \frac{(x+h) + \sqrt{x+h} - (x + \sqrt{x})}{h}

    I can't seem to get this one going. I can easily simplify it down to:

    \displaystyle \lim_{h\to 0 } \frac{h + \sqrt{x+h} - \sqrt{x}}{h}

    But then I don't know where to take it. There is no easy conjugate to multiply by, and I don't see any way to factor the numerator. Can anybody give me a hint about what operation to try on this thing to move it along?

    Thanks.
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  2. #2
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    Quote Originally Posted by joatmon View Post
    Find f'(x) for the function f(x) = x + \sqrt{x}

    This is Calc 1, so we have to do this the long way for a little while longer, so I have to use the formula

    f'(x) = \displaystyle \lim_{h\to 0 } \frac{(x+h) + \sqrt{x+h} - (x + \sqrt{x})}{h}

    I can't seem to get this one going. I can easily simplify it down to:

    \displaystyle \lim_{h\to 0 } \frac{h + \sqrt{x+h} - \sqrt{x}}{h}

    But then I don't know where to take it. There is no easy conjugate to multiply by, and I don't see any way to factor the numerator. Can anybody give me a hint about what operation to try on this thing to move it along?

    Thanks.
    \displaystyle \lim_{h\to 0 } \frac{h + \sqrt{x+h} - \sqrt{x}}{h}=1+ \lim_{h\to 0 } \frac{ \sqrt{x+h} - \sqrt{x}}{h}

    Now:

    \displaystyle \lim_{h\to 0 } \frac{ \sqrt{x+h} - \sqrt{x}}{h}=\lim_{h\to 0 } \frac{ (\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}

    ....

    CB
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  3. #3
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    Sure. That will work. Thanks a lot. I appreciate it.
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