Find the derivative

  • Thread starter shwanky
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  • #1
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Homework Statement


[tex]f(x) = x - \sqrt{x}[/tex]


Homework Equations


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


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.
 
Last edited:

Answers and Replies

  • #2
529
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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.
 
  • #3
529
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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)
 
  • #4
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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.
 
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  • #5
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Using the limit definition is so tedious :(
 
  • #6
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Got it...working on the tex
 
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  • #7
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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.
 
  • #8
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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 [tex]\sqrt{x+h}+\sqrt{x}\\[/tex]

edit: yay my comp is working lets give this another try

[tex]
\[\lim_{h \to 0}\frac{x+h+\sqrt{x+h}-x-\sqrt{x}}{h}\]
& =\]\lim_{h \to 0}\frac{h+\sqrt{x+h}-\sqrt{x}}{h}\]
& =\[\lim_{h \to 0}\frac{h(\sqrt{x+h}+\sqrt{x})+x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]
& =\[\lim_{h \to 0}\frac{h(\sqrt{x+h}+\sqrt{x})+h}{h(\sqrt{x+h}+\sqrt{x})}\]
& =\[\lim_{h \to 0}1+\frac{1}{\sqrt{x+h}+\sqrt{x}}\]
& =\[1+\frac{1}{2\sqrt{x}}\]
[/tex]

Damn I suck at this.
 
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  • #9
43
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ok, I'll have to practice that one. :) thanks
 
  • #10
11
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U can also use the addition rules of limits to calculate derivatives for x and sqrt x separately and then add them.
 

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