1. ## Find the derivative

Darn professors....wish mine explained things...

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

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.

2. Think what can you multiply the expression by that will lead to the expression to a difference of two squares?

3. 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.

4. 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.

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

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

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.
$\lim_{x \rightarrow a} \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a} = \lim_{x \rightarrow a} \left( \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a}\right) \cdot \left( \frac{ \sqrt{3x + 1} + \sqrt{3a + 1} }{\sqrt{3x + 1} + \sqrt{3a + 1}} \right) = \, ....$

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