Limit definition of a derivative

[crybllrd]

Homework Statement

I had my second exam last week for my Calculus I course. I did alright, but we are supposed to correct them and bring them back for a quiz grade. However, I wasn't sure how to do this one on the test, and did not magically figure it out since then :)

Find the derivative of $$f(x)=\sqrt{3x+1}$$

Homework Equations

$$f'(x)=\frac{f(x+h)-f(x)}{h}$$

The Attempt at a Solution

$$f'(x)=\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}$$


I have a hard time with these limit definitions of a derivative when it involves a square root. I am thinking that I need to use a conjugate, but I am really not sure how.

 

[Dick]

Multiply numerator and denominator by sqrt(3(x+h)+1)+sqrt(3x+1), right? That's the usual conjugate thing.

 

[crybllrd]

$$\frac{\sqrt{3(x+h)+1}-\sqrt{3x+1}}{h}\cdot \frac{\sqrt{3(x+h)+1}+\sqrt{3x+1}}{\sqrt{3(x+h)+1}+\sqrt{3x+1}}$$

I am not sure about the denominator on this next step:

$$\frac{3h}{h(\sqrt{3(x+h)+1}+\sqrt{3x+1})}$$

I know the answer should be

$$\frac{3}{2\sqrt{3x+1}}$$

How do I get to that?

 

[HallsofIvy]

There should be an obvious step to simplify
$$\frac{3h}{h(\sqrt{3(x+h)+ 1}+ \sqrt{3x+1})}$$

and after that you can take the limit by just taking h= 0.

 

[crybllrd]

OK thanks, I got it figured out from here. I just needed to divide top and bottom by h, then distribute under the radical, the remaining h equals zero, and I ended up with the correct answer.
Thanks again!