Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question
please read how to use Mathjax:… – Applied mathematician Jan 14 '13 at 18:58
Sorry, I meant the previous expression is above the denominator of $h$ – Rebecca Korbal Jan 14 '13 at 19:15

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 $\frac1{\sqrt{x+h}}+\frac1{\sqrt{x}}$.

share|cite|improve this answer

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 $1/\sqrt{x}$.

share|cite|improve this answer

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.

share|cite|improve this answer
It would seem from the question that it arises from finding the derivative of $\frac1{\sqrt x}$. Furthermore, $$\left(\frac1{\sqrt x}\right)'=-\frac1{2\sqrt{x}^3}$$ – robjohn Jan 14 '13 at 19:20
@robjohn: Opppsss. Wowwww. Thanks for noting me that. – Babak S. Jan 14 '13 at 19:24
oops...almost! +1 – amWhy Feb 17 '13 at 0:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.