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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.
  2. jcsd
  3. Jul 10, 2012 #2


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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


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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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