# Find the derivative

• Feb 5th 2009, 07:23 PM
TastyBeverage
Find the derivative
Darn professors....wish mine explained things...

How do I find f(a) for$\displaystyle f(x) = (3x +1)^{1/2}$

Thats the square root of 3x+1, double thanks to someone who shows me how to put that in there.

I have only seen f(x) - f(a)/x-a and f(a + h) - f(a)/h, so no dx or whatever that stuff is please.
• Feb 5th 2009, 07:49 PM
Oiram
Think what can you multiply the expression by that will lead to the expression to a difference of two squares?
• Feb 5th 2009, 07:56 PM
TastyBeverage
Quote:

Originally Posted by Oiram
Think what can you multiply the expression by that will lead to the expression to a difference of two squares?

what?! assume im brain dead cause i basically am.
• Feb 5th 2009, 07:58 PM
Oiram
Check this out:
Difference of two squares - Wikipedia, the free encyclopedia
Although the expression may be under the square root operation you may still use the technique found in the link above.
• Feb 5th 2009, 09:07 PM
mr fantastic
Quote:

Originally Posted by TastyBeverage
Darn professors....wish mine explained things...

How do I find f(a) for$\displaystyle f(x) = (3x +1)^{1/2}$

Thats the square root of 3x+1, double thanks to someone who shows me how to put that in there.

I have only seen f(x) - f(a)/x-a and f(a + h) - f(a)/h, so no dx or whatever that stuff is please.

$\displaystyle \lim_{x \rightarrow a} \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a} = \lim_{x \rightarrow a} \left( \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a}\right) \cdot \left( \frac{ \sqrt{3x + 1} + \sqrt{3a + 1} }{\sqrt{3x + 1} + \sqrt{3a + 1}} \right) = \, ....$
• Feb 5th 2009, 09:45 PM
TastyBeverage
Quote:

Originally Posted by mr fantastic
$\displaystyle \lim_{x \rightarrow a} \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a} = \lim_{x \rightarrow a} \left( \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a}\right) \cdot \left( \frac{ \sqrt{3x + 1} + \sqrt{3a + 1} }{\sqrt{3x + 1} + \sqrt{3a + 1}} \right) = \, ....$

ahhh....rationalize. Thanks!