# Derivative problem...how did they get this answer?

• Jan 19th 2009, 10:13 AM
janedoe
Derivative problem...how did they get this answer?
http://i43.tinypic.com/2j11mgy.jpg --> problem is here

I only would like to know how they got that from the derivative equation. Can you please show me step by step how you'd go about doing it, using that formula?

Thanks!!
• Jan 19th 2009, 10:37 AM
galactus
They did this:

$\displaystyle \lim_{h\to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$

How, work the algebra magic, take the limit as h approaches 0 and you have it. The derivative of $\displaystyle \frac{1}{\sqrt{x}}$

Note: I used h instead of $\displaystyle {\Delta}x$. Same thing
• Jan 19th 2009, 10:43 AM
janedoe
I got that too but I don't understand what comes next..can you please show me the next steps?
• Jan 19th 2009, 11:02 AM
Jester
Quote:

Originally Posted by galactus
They did this:

$\displaystyle \lim_{h\to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$

How, work the algebra magic, take the limit as h approaches 0 and you have it. The derivative of $\displaystyle \frac{1}{\sqrt{x}}$

Note: I used h instead of $\displaystyle {\Delta}x$. Same thing

Next hint, simplify what Galactus has as

$\displaystyle \lim_{h\to 0}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x+h}\sqrt{x}}$ then rationalize.
• Jan 19th 2009, 11:25 AM
janedoe
I did that and got your answer but now I'm having trouble rationalizing...could you show me how to go about that with this problem?
• Jan 19th 2009, 12:31 PM
Jester
Quote:

Originally Posted by janedoe
I did that and got your answer but now I'm having trouble rationalizing...could you show me how to go about that with this problem?

$\displaystyle \lim_{h\to 0}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x+h}\sqrt{x}}$$\displaystyle = \lim_{h\to 0}\frac{\sqrt{x}-\sqrt{x+h}} {h\sqrt{x+h}\sqrt{x}}\cdot \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}$

Expand only the numerator and things in the top will cancel, then cancel the $\displaystyle h$ in the bottom, then take the limit. :)