# How do I simplify this fractional expression?

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

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please read how to use Mathjax: meta.math.stackexchange.com/questions/5020/… – 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 $\frac1{\sqrt{x+h}}+\frac1{\sqrt{x}}$.

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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 $1/\sqrt{x}$.

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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.

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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