$\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}$ and these numbers are all above the denominator of $h$.

Can someone please help me to understand how to simplify this expression?


3 Answers 3


I'm not sure if I interpreted your math correctly, but in general if you have something like

$\frac1{\sqrt{a}-\sqrt{b}}$, multiply the numerator and denominator by $\sqrt{a} + \sqrt{b}$

This will cancel out the square root operators from the denominator.

In your case, if I interpreted correctly, multiply the numerator and denominator by $\frac1{\sqrt{x+h}}+\frac1{\sqrt{x}}$.


Combine and then simplify the numerator:

$$\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}} = -\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}\sqrt{x}} $$

Use the fact that

$$ \sqrt{x+h}-\sqrt{x} = \frac{h}{\sqrt{x+h}+\sqrt{x}} $$

to get

$$\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}} = -\frac{h}{(\sqrt{x+h}+\sqrt{x})\sqrt{x+h}\sqrt{x} } $$

I imagine you need this to compute the derivative of $1/\sqrt{x}$.


If you knew the derivation, I would tell you that $$\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}\sim h\left(\frac{1}{\sqrt{x}}\right)'=h\frac{1}{-2\sqrt{x^3}}$$ when $h$ is so small.

  • $\begingroup$ It would seem from the question that it arises from finding the derivative of $\frac1{\sqrt x}$. Furthermore, $$\left(\frac1{\sqrt x}\right)'=-\frac1{2\sqrt{x}^3}$$ $\endgroup$
    – robjohn
    Jan 14, 2013 at 19:20
  • $\begingroup$ @robjohn: Opppsss. Wowwww. Thanks for noting me that. $\endgroup$
    – Mikasa
    Jan 14, 2013 at 19:24
  • $\begingroup$ oops...almost! +1 $\endgroup$
    – amWhy
    Feb 17, 2013 at 0:06

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