anonymous
  • anonymous
can someone help me find the derivative to this equation 2xsqrtx^2+1
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[2x \sqrt{x^2+1}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
lemme do that again ... i just realized i didn't take the square root..

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
|dw:1333008783149:dw|
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1333008899711:dw|
anonymous
  • anonymous
factor out the common factor of 2(x^2+1)^(-1/2)
anonymous
  • anonymous
|dw:1333009455298:dw|
anonymous
  • anonymous
sorry i dont know what you did this is where am at 2[x(2x)/2sqrtx^2+1+sqrt x^2+1
anonymous
  • anonymous
Plz see the attachment
1 Attachment
anonymous
  • anonymous
Is the photo clear? If not, I can upload another one
anonymous
  • anonymous
i can't make out what you wrote for your derivative (@ goku3)
anonymous
  • anonymous
would mind uploading another pic @CoCoTsoi
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I agree with @CoCoTsoi 's answer.
anonymous
  • anonymous
It is beauteeeful.
anonymous
  • anonymous
i get lost between step to and three
anonymous
  • anonymous
Chain rule
anonymous
  • anonymous
why is there two 2x on top
anonymous
  • anonymous
And product rule... http://tutorial.math.lamar.edu/Classes/CalcI/ProductQuotientRule.aspx
anonymous
  • anonymous
one 2x is the original one. another is by diff the equation inside the sqrt
anonymous
  • anonymous
So say if you have f*g. If you use the product rule, then you'll do: f'*g + f*g'.
anonymous
  • anonymous
sorry if you couldn't see the tick mark next to the f.
anonymous
  • anonymous
its cool how you get rid of the sqrt on top
anonymous
  • anonymous
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)
anonymous
  • anonymous
lol... wants not want's. I type too fast.
anonymous
  • anonymous
multiply numerator and denominator with sqrt(x^2+1)
anonymous
  • anonymous
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
anonymous
  • anonymous
We all tend to get lost in the little steps, believe me I've been there. Coco did a great job writing that out.
anonymous
  • anonymous
Coz I was lazy for typing it on keyboard so I chose to write it on paper :D
anonymous
  • anonymous
paper is way faster and thank u guys for being so patient with me
anonymous
  • anonymous
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.
anonymous
  • anonymous
Thanks for giving a chance for us to learn
anonymous
  • anonymous
yeah i will i just have to take my on these problems and not rush them
anonymous
  • anonymous
Check out paul's online notes if you haven't already, they are great :-)
anonymous
  • anonymous
i have them bookmarked actually

Looking for something else?

Not the answer you are looking for? Search for more explanations.