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goku3 Group Title

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

  • 2 years ago
  • 2 years ago

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  1. goku3 Group Title
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    \[2x \sqrt{x^2+1}\]

    • 2 years ago
  2. goku3 Group Title
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    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

    • 2 years ago
  3. dpaInc Group Title
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    lemme do that again ... i just realized i didn't take the square root..

    • 2 years ago
  4. dpaInc Group Title
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    |dw:1333008783149:dw|

    • 2 years ago
  5. goku3 Group Title
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    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

    • 2 years ago
  6. dpaInc Group Title
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    |dw:1333008899711:dw|

    • 2 years ago
  7. dpaInc Group Title
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    factor out the common factor of 2(x^2+1)^(-1/2)

    • 2 years ago
  8. dpaInc Group Title
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    |dw:1333009455298:dw|

    • 2 years ago
  9. goku3 Group Title
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    sorry i dont know what you did this is where am at 2[x(2x)/2sqrtx^2+1+sqrt x^2+1

    • 2 years ago
  10. CoCoTsoi Group Title
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    Plz see the attachment

    • 2 years ago
    1 Attachment
  11. CoCoTsoi Group Title
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    Is the photo clear? If not, I can upload another one

    • 2 years ago
  12. dpaInc Group Title
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    i can't make out what you wrote for your derivative (@ goku3)

    • 2 years ago
  13. goku3 Group Title
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    would mind uploading another pic @CoCoTsoi

    • 2 years ago
  14. CoCoTsoi Group Title
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    • 2 years ago
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  15. brinethery Group Title
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    I agree with @CoCoTsoi 's answer.

    • 2 years ago
  16. brinethery Group Title
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    It is beauteeeful.

    • 2 years ago
  17. goku3 Group Title
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    i get lost between step to and three

    • 2 years ago
  18. brinethery Group Title
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    Chain rule

    • 2 years ago
  19. goku3 Group Title
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    why is there two 2x on top

    • 2 years ago
  20. brinethery Group Title
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    And product rule... http://tutorial.math.lamar.edu/Classes/CalcI/ProductQuotientRule.aspx

    • 2 years ago
  21. CoCoTsoi Group Title
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    one 2x is the original one. another is by diff the equation inside the sqrt

    • 2 years ago
  22. brinethery Group Title
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    So say if you have f*g. If you use the product rule, then you'll do: f'*g + f*g'.

    • 2 years ago
  23. brinethery Group Title
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    sorry if you couldn't see the tick mark next to the f.

    • 2 years ago
  24. goku3 Group Title
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    its cool how you get rid of the sqrt on top

    • 2 years ago
  25. brinethery Group Title
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    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)

    • 2 years ago
  26. brinethery Group Title
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    lol... wants not want's. I type too fast.

    • 2 years ago
  27. CoCoTsoi Group Title
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    multiply numerator and denominator with sqrt(x^2+1)

    • 2 years ago
  28. goku3 Group Title
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    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

    • 2 years ago
  29. brinethery Group Title
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    We all tend to get lost in the little steps, believe me I've been there. Coco did a great job writing that out.

    • 2 years ago
  30. CoCoTsoi Group Title
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    Coz I was lazy for typing it on keyboard so I chose to write it on paper :D

    • 2 years ago
  31. goku3 Group Title
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    paper is way faster and thank u guys for being so patient with me

    • 2 years ago
  32. brinethery Group Title
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    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.

    • 2 years ago
  33. CoCoTsoi Group Title
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    Thanks for giving a chance for us to learn

    • 2 years ago
  34. goku3 Group Title
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    yeah i will i just have to take my on these problems and not rush them

    • 2 years ago
  35. brinethery Group Title
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    Check out paul's online notes if you haven't already, they are great :-)

    • 2 years ago
  36. goku3 Group Title
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    i have them bookmarked actually

    • 2 years ago
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