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\[2x \sqrt{x^2+1}\]

lemme do that again ... i just realized i didn't take the square root..

|dw:1333008783149:dw|

4x^2/2sqrtx^2+1 +x^2+1 and i dont know how to simplify this to get 2(2x^2+1)/sqrtx^2+1

|dw:1333008899711:dw|

factor out the common factor of 2(x^2+1)^(-1/2)

|dw:1333009455298:dw|

sorry i dont know what you did this is where am at 2[x(2x)/2sqrtx^2+1+sqrt x^2+1

Plz see the attachment

Is the photo clear? If not, I can upload another one

i can't make out what you wrote for your derivative (@ goku3)

It is beauteeeful.

i get lost between step to and three

Chain rule

why is there two 2x on top

And product rule...
http://tutorial.math.lamar.edu/Classes/CalcI/ProductQuotientRule.aspx

one 2x is the original one. another is by diff the equation inside the sqrt

So say if you have f*g. If you use the product rule, then you'll do:
f'*g + f*g'.

sorry if you couldn't see the tick mark next to the f.

its cool how you get rid of the sqrt on top

lol... wants not want's. I type too fast.

multiply numerator and denominator with sqrt(x^2+1)

Coz I was lazy for typing it on keyboard so I chose to write it on paper :D

paper is way faster and thank u guys for being so patient with me

Thanks for giving a chance for us to learn

yeah i will i just have to take my on these problems and not rush them

Check out paul's online notes if you haven't already, they are great :-)

i have them bookmarked actually