square root derivative using limit definition of derivative

K

Kristy

Junior Member
Joined
Jun 26, 2007
Messages
57
Calc 1 summer college class: #26

The directions are: Find f'[x] given f[x] and they want us to do it using definition of a limit. Also state domain of function and its derivative.

[There were alot of other problems and I got all of the other ones so I know the basic idea I just can't figure out how to do it with squre roots!]

(limit as h ->0 of f[x+h]-f[x] / h

I know the answer should be 1/ 2 square root x

I got as far as substituting in the function
f[x]=x+ sqrt x
f[x+h]=x+h + sqrt [x+h]

Then I substituted those in the definition of the limit above, there was an x-x that cancellled in the numerator, then I separated the h by first separating it into 2 fractions, then the limit law that the limit of sums is thee sums of limits. And then that's where I got stuck. I tried directly plugging things in, tried dividing each term by x, and by x2. I tried separating out all of the square roots.

here is my last step that seems like I'm getting anywhere.
f'[x]=1+ limit as h->0 of
numerator sqrt[x+h] + sqrt [x]
denominator h

I did a google search of square root limit, definition of derivative, and didn't come up with anything that helpful. The only thing I think might be possible is "conjugate" to rationalize. So Im going to try that but I can't think of anything else!

(Sorry if it is kind of hard to understand this, I am not sure how to make it appear like it does when I write it out. It makes it more confusing for me too.) Also I do not know why the time shows wrong when I click on preview. Wed Jun 27, 2007 12:47 am is what it showed and it is really only almost 9pm on the 26th
 
The trick is to multiply by the conjugate of the numerator

\(\displaystyle \L\\\frac{d}{dx}[\sqrt{x}]\):

\(\displaystyle \L\\\lim_{h\to\0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{\not{h}}{\not{h}(\sqrt{x+h}+\sqrt{x})}\)

=\(\displaystyle \L\\\lim_{h\to\0}\frac{1}{\sqrt{x+h}+\sqrt{x}}\)

=\(\displaystyle \L\\\frac{1}{2\sqrt{x}}\)

As for the time, go to your profile and set the tme to GMT-4 hrs if you're on the East Coast of the US
It's currently on GMT(Greenwich Mean Time).
 
Also I got that the domain for the function and derivative are
[0,inf)


Thanks for the information! (On how to solve it and that fixed the time issue.)


Also, how do you get it to look like that (almost like handwritten)
 
I am sorry, I had a typo in the first step. I fixed it.

0 is not in the domain of the derivative. See why?.

As for the style, that's LaTex. Click on 'quote' at the upper right hand corner of my post to see the code I used to make it display that way.
 
No, I don't understand, I thought 0 was in the domain. It seems like
x + square root x
would be 0+0=0

and for the derivative, okay, I guess I can see why it would not be zero since 1/2*sqrtx = 1/2*0 is undefined.

Also, I did do the quote and that looks VERY complicated, but interesting!
 
Not in the derivative. If you put 0 in the derivative you have division by 0. As you should know, that's a big no-no.

\(\displaystyle \L\\\frac{1}{2\sqrt{0}}=undefined\)
 
You did not post your original function. I just guessed it was \(\displaystyle \L\\\sqrt{x}\). Was it something else?.

EDIT: I'm sorry, you did. I did't see it. That just makes it \(\displaystyle \L\\\frac{1}{2\sqrt{x}}+1\). No big deal.
 
Take a good, close look at that derivative function.

\(\displaystyle \L\;f'(x)\;=\;\frac{1}{2\sqrt{x}}\)

Is zero (0) in or out?
 
Re: square root derivative using limit definition of derivat

Kristy said:
I got as far as substituting in the function
f[x]=x+ sqrt x
f[x+h]=x+h + sqrt [x+h]
....
Click to expand...


Sorry, I guess I should have made it more clear that that was f[x] and the original function.
 
tkhunny said:
Take a good, close look at that derivative function.

\(\displaystyle \L\;f'(x)\;=\;\frac{1}{2\sqrt{x}}\) +1
Is zero (0) in or out?
Click to expand...

I think zero is still not included. Because you would sort of get infinity plus 1 which would still be undefined. Right?
 
Division by zero is never allowed. Also, think of the limit as x goes to 0 as being infinity. The actual expression 1/0 is meaningless and it is not true that it is EQUAL to infinity.
 
You must log in or register to reply here.
Top