Solving for the Derivative of a Square Root Function

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derivative problem...

Homework Statement



f(x)=x-2/x^2+1 Square root

f(x)=lim f(x-h) - f(x)/h
h>0

f(x)= lim x+h-2/(x+h)^2+1 Square root - x-2/x^2+1 Square root
h>0 /h



Homework Equations





The Attempt at a Solution



I do have the answer... 2x+1/(x^2+1)^3/2

I know the first step would be to cross multiply the ones in the 3rd part. I get lost after that


man, I hope all that's understandable. hopefully someone can teach me how to do square roots and other stuff

just in case, the square roots go in x^2+1, (x+h)^2+1, and x^2+1 which are all denominators. and the h divides everything in the third part
 
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Dude487 said:

Homework Statement



f(x)=x-2/x^2+1 Square root
This is so ambiguous I can't tell what the function is. You absolutely need parentheses here. Also, exactly what expression is inside the radical?

I'll take a wild stab at what I think you might mean:
f(x) = \sqrt{\frac{x -2}{x^2 + 1}}

It's a safe bet that you don't know LaTeX scripting, so function formula I wrote above should be written something like this so that people can understand what you're trying to do.
f(x) = sqrt[(x - 2)/(x^2 + 1)]


Dude487 said:
f(x)=lim f(x-h) - f(x)/h
h>0
The expression in the limit should be written as [f(x + h) - f(x)]/h
Dude487 said:
f(x)= lim x+h-2/(x+h)^2+1 Square root - x-2/x^2+1 Square root
h>0 /h



Homework Equations





The Attempt at a Solution



I do have the answer... 2x+1/(x^2+1)^3/2
Hallelujah, do I see a pair of parentheses? It would be better with another pair around the 2x + 1 part.
Dude487 said:
I know the first step would be to cross multiply the ones in the 3rd part. I get lost after that
Cross multiply? Why would you do that?
Dude487 said:
man, I hope all that's understandable. hopefully someone can teach me how to do square roots and other stuff

just in case, the square roots go in x^2+1, (x+h)^2+1, and x^2+1 which are all denominators. and the h divides everything in the third part
 


This is a fairly tricky problem if you have to do it by using the definition of the derivative. If you are permitted to use some other technique, I would recommend doing it that way.

Here's to get you started:
f&#039;(x) = \lim_{h \rightarrow 0} 1/h [f(x + h) - f(x)]<br /> = \lim_{h \rightarrow 0} 1/h [ \sqrt{\frac{x + h - 2}{(x + h)^2 + 1}} - \sqrt{\frac{x -2}{x^2 + 1}}]<br /> = \lim_{h \rightarrow 0} 1/h [ \frac{\sqrt{x + h - 2}}{\sqrt{(x + h)^2 + 1}} - \frac{\sqrt{x - 2}}{\sqrt{x^2 + 1}}]<br />

Now it starts to get really ugly. You'll have to get a common denominator, and then you'll need to multiply the whole expression inside the brackets by 1, in the form of the sum of two radicals over itself. Then you can multiply by 1/h, and finally take the limit as h approaches zero.
 
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