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can someone help me find the derivative to this equation 2xsqrtx^2+1

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\[2x \sqrt{x^2+1}\]
actually i got all to the end of the i just can figure out how to simlify the damn thing to match the books answer
lemme do that again ... i just realized i didn't take the square root..

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Other answers:

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
factor out the common factor of 2(x^2+1)^(-1/2)
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
1 Attachment
Is the photo clear? If not, I can upload another one
i can't make out what you wrote for your derivative (@ goku3)
would mind uploading another pic @CoCoTsoi
1 Attachment
I agree with @CoCoTsoi 's answer.
It is beauteeeful.
i get lost between step to and three
Chain rule
why is there two 2x on top
And product rule...
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
Okay so going from the 3rd to the 4th line: There's 2sqrt(x^2+1) He want's to get it so it has the same denominator as the other term. The den of the other term is sqrt(x^2+1). So with the left term, you have to multiply the numerator and denominator by sqrt(x^2+1). Then you end up with (2sqrt(x^2+1)*sqrt(x^2+1))/sqrt(x^2+1) = (2(x^2+1))/sqrt(x^2+1)
lol... wants not want's. I type too fast.
multiply numerator and denominator with sqrt(x^2+1)
thank you guys so much every time i tried to do it i forgot to add one of the 2x and messed my entire thing up
We all tend to get lost in the little steps, believe me I've been there. Coco did a great job writing that out.
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
So goku, remember with this problem, you have to use the product rule and then the chain rule since you have stuff inside of that square root.
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

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