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The derivative of 1/sqrt(x)

  1. Jan 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of [tex]\frac{1}{\sqrt{x}}[/tex] using the lim definition.

    2. Relevant equations
    f(x)'=[tex]\frac{f(x+h)-f(x)}{h}[/tex]

    3. The attempt at a solution
    Keep in mind that everything bellow is for the lim as h approaches 0.

    [tex]\frac{1}{\sqrt{x}}[/tex]

    [tex]\Downarrow[/tex]

    [tex]
    \frac{
    \frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}
    }
    {h}
    [/tex]

    [tex]\Downarrow[/tex]

    (I multiply both nominator and denominator with conjugate)

    [tex]
    \frac
    {
    \frac{1}{x+h}-\frac{1}{x}
    }
    {
    \frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}
    }
    [/tex]

    After this I am totally lost..
     
  2. jcsd
  3. Jan 6, 2011 #2

    Dick

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    Science Advisor
    Homework Helper

    Combine numerator into a single fraction. See if you get an h you can cancel with the h in the denominator.
     
  4. Jan 6, 2011 #3
    You could also use the definition...

    [tex]f'(x)= \lim_{ x \to a} \frac{f(x)- f(a)}{x-a} [/tex].
     
  5. Jan 6, 2011 #4
    Thaaaank you! Problem solved! :)
    Did the same to denominator and then combined the two franctions into one.
     
  6. Jan 6, 2011 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Glad you got it solved. As a check, remember that you can write [itex]\frac{1}{\sqrt{x}}[/itex] as [itex]x^{-1/2}[/itex] and use the power rule.
     
  7. Jan 6, 2011 #6

    SammyS

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    Staff Emeritus
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    Homework Helper
    Gold Member

    This is perfectly fine - up to this point.
    Continuing on:

    [tex]\displaystyle =\frac
    { \displaystyle \frac{x-(x+h)}{(x+h)x}
    }
    { \displaystyle \frac{h}{\ \sqrt{x+h}}+\frac{h}{\sqrt{x}\ \ }
    } [/tex]

    [tex]\displaystyle =\frac
    { \displaystyle \frac{-h}{(x+h)(x)}\ \cdot\ \displaystyle \frac{1}{h}
    }
    { \displaystyle \left(\frac{h}{\sqrt{x+h}}+\frac{h}{\sqrt{x}}\right) \ \cdot\ \displaystyle \frac{1}{h}}
    }
    [/tex]

    [tex]\displaystyle =\frac
    { \displaystyle \frac{-1}{(x+h)(x)}
    }
    { \displaystyle \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}
    }
    [/tex]

    Then,
    [tex]\displaystyle f'(x)= \lim_{h\to 0} \ \
    \frac
    { \displaystyle \frac{-1}{(x+h)(x)}
    }
    { \displaystyle \frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}
    }
    [/tex]
     
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