2
$\begingroup$

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

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

$\endgroup$
5
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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 $1/\sqrt{x}$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '13 at 19:20
  • $\begingroup$ @robjohn: Opppsss. Wowwww. Thanks for noting me that. $\endgroup$ – mrs Jan 14 '13 at 19:24
  • $\begingroup$ oops...almost! +1 $\endgroup$ – Namaste Feb 17 '13 at 0:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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