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Derivative of function with radicals

  1. Jul 10, 2012 #1
    I've been having trouble figuring out how to find the derivative of f(x) = x + √x

    The farthest I got was:

    [(x+h) + √(x+h) - (x+√x)]/h =

    [h + √(x+h) - √x] / h

    I got stuck here because i'm not sure how to cancel out h in numerator and denominator (if i can even do that at this stage) or multiply the whole equation by the conjugate. For some reason i'm stumped on what seems like simple algebra.
     
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  3. Jul 10, 2012 #2

    HallsofIvy

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    That is, of course, [itex]1+ \frac{\sqrt{x+h}- \sqrt{x}}{h}[/itex]. The "1" is of course, the derivative of "x". All that is left is the last fraction. To do that "rationalize" the numerator: multiply both numerator and denominator by [itex]\sqrt{x+h}+ \sqrt{x}[/itex] to get
    [tex]\frac{\sqrt{x+h}- \sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}[/tex]
    [tex]= \frac{(x+ h)- x}{h(\sqrt{x+h}+\sqrt{x})}= \frac{1}{\sqrt{x+h}+ \sqrt{x}}[/tex]

    Now, it's easy to take the limit as h goes to 0.
     
  4. Jul 10, 2012 #3
    I see, thank you for the help! the "1 + (fraction)" was what I missed.

    Does this mean my second step [h + √(x+h) - √x] / h was wrong?

    I'm still not sure how to arrive at your first step, the "1 + (fraction)". However, everything else makes sense now!
     
  5. Jul 10, 2012 #4

    eumyang

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    No your second step isn't wrong. HallsofIvy's first step comes from your second step:
    [tex]\frac{h + \sqrt{x+h}- \sqrt{x}}{h}[/tex]
    [tex]= \frac{h}{h} + \frac{\sqrt{x+h}- \sqrt{x}}{h}[/tex]
    [tex]= 1 + \frac{\sqrt{x+h}- \sqrt{x}}{h}[/tex]
     
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