Find Derivative of f(x) = x - \sqrt{x}

[shwanky]

Homework Statement

f(x) = x - \sqrt{x}

Homework Equations

f'(x) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}

The Attempt at a Solution

I'm not exactly sure how to solve this problem. I plugged the problem into the equation and got stuck... I have a class right now, so I'll have to post my work later.

 

[christianjb]

It's not easy to find the derivative of sqrt(x) in that way. (You can't use a Taylor expansion- because that makes use of the answer.)

I think you have to do a binomial expansion of sqrt(x)
http://www.rism.com/Trig/binomial.htm

There may be a very easy solution- but I don't see it.

 

[christianjb]

Duh, wait a minute...

if g(x)=x^1/2, then g(x)^2=x. Now take derivatives of both sides- which can be proved to give:

2g(x)dg(x)/dx=1

dg(x)/dx=1./2g(x)=1/2 (x^-1/2)

So a method which is a little easier involves proving
d[g(x)^2]/dx=2g(x)dg(x)/dx, for any fn. g(x)

 

[wombat7373]

I just tried it and got a wrong answer. Your goal is the get everything in the numerator so that it has an h in it. That way u can cancel out the h on the bottom.

 

[wombat7373]

Using the limit definition is so tedious :(

 

[wombat7373]

Got it...working on the tex

 

[shwanky]

Well, I read one chapter a head so I figured out how to do it using the power rule.. :-/ it's a way easier that way.

 

[wombat7373]

My comp. is being stupid so I can't write the whole thing =\ I had a test once where we had to do a lot of those derivatives without the shortcuts. U can solve it that way by multiplying both the numerator and the denominator by \sqrt{x+h}+\sqrt{x}\\

edit: yay my comp is working let's give this another try

<br /> \[\lim_{h \to 0}\frac{x+h+\sqrt{x+h}-x-\sqrt{x}}{h}\]<br /> &amp; =\]\lim_{h \to 0}\frac{h+\sqrt{x+h}-\sqrt{x}}{h}\]<br /> &amp; =\[\lim_{h \to 0}\frac{h(\sqrt{x+h}+\sqrt{x})+x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]<br /> &amp; =\[\lim_{h \to 0}\frac{h(\sqrt{x+h}+\sqrt{x})+h}{h(\sqrt{x+h}+\sqrt{x})}\]<br /> &amp; =\[\lim_{h \to 0}1+\frac{1}{\sqrt{x+h}+\sqrt{x}}\]<br /> &amp; =\[1+\frac{1}{2\sqrt{x}}\]<br />

Damn I suck at this.

 

[shwanky]

ok, I'll have to practice that one. :) thanks

 

[wombat7373]

U can also use the addition rules of limits to calculate derivatives for x and sqrt x separately and then add them.