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sandy524 Group Title

find the derivative of f(x)=sqrt (x^2+2x)

  • 2 years ago
  • 2 years ago

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  1. sandy524 Group Title
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    \[f(x)=\sqrt{x^2+2}\]

    • 2 years ago
  2. swissgirl Group Title
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    \( f(x)=(x^2+2)^{1 \over 2}\)

    • 2 years ago
  3. sandy524 Group Title
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    thats not one of the choices

    • 2 years ago
  4. swissgirl Group Title
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    noooooooo its easier to solve in this format

    • 2 years ago
  5. sandy524 Group Title
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    huh

    • 2 years ago
  6. sandy524 Group Title
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    #4

    • 2 years ago
  7. rajdoshi Group Title
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    \[1\div2\sqrt{x ^{2}+2x}\times(2x+2)\]

    • 2 years ago
  8. swissgirl Group Title
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    \( f(x)=(x^2+2)^{1 \over 2}\) is the same as \( f(x)= \sqrt{x^2+2}\)

    • 2 years ago
  9. zepdrix Group Title
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    Sandy, it's easier to differentiate a root if you rewrite it as a fractional exponent :) She was just rewriting the problem a little nicer for you, not the solution. :D

    • 2 years ago
  10. sandy524 Group Title
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    i understand but the assignment is keeping it that way so im getting confused thats all

    • 2 years ago
  11. zepdrix Group Title
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    it might be a good idea to switch to a fractional exponent, do the differentiation, then as your last step, put whatever fractional exponent you have back into root form. UNLESS you remember the derivative of sqrt(x), then you skip doing that step ^^

    • 2 years ago
  12. zepdrix Group Title
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    swiss is typing up a storm :P she prolly got something for you lol

    • 2 years ago
  13. sandy524 Group Title
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    lol

    • 2 years ago
  14. zepdrix Group Title
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    she might just be trolling us, it's hard to tell at this point -_- lol

    • 2 years ago
  15. swissgirl Group Title
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    \(f(x)=(x^2+2)^{1 \over 2}\) \( {1 \over 2}(x^2+2)^{({1 \over 2}-1)}*2x\) \( {1 \over 2}(x^2+2)^{-1 \over 2}*2x\) \(\large { 1 \over 2\sqrt{x^2+2}}*2x\) \(\large {2x \over 2(x^2+2)}\) \(\large {x \over (x^2+2)}\)

    • 2 years ago
  16. swissgirl Group Title
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    okkk i had to post loil but if you are still interested I can explain what i did

    • 2 years ago
  17. zepdrix Group Title
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    Woops! Your sqrt disappeared on the last couple steps there! <:o

    • 2 years ago
  18. swissgirl Group Title
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    @zepdrix was getting too curious

    • 2 years ago
  19. swissgirl Group Title
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    uuugghhhhhhhhhh

    • 2 years ago
  20. zepdrix Group Title
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    XD

    • 2 years ago
  21. swissgirl Group Title
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    i hate latex it never listens

    • 2 years ago
  22. swissgirl Group Title
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    ok one more time at it

    • 2 years ago
  23. sandy524 Group Title
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    :(

    • 2 years ago
  24. sandy524 Group Title
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    ok thanks

    • 2 years ago
  25. sandy524 Group Title
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    you guys rock

    • 2 years ago
  26. zepdrix Group Title
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    Do the first 3 steps make sense though sandy? c: those she did correctly.

    • 2 years ago
  27. swissgirl Group Title
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    ok lets explain the first 3 steps that last 2 steps are basic simplification neways

    • 2 years ago
  28. sandy524 Group Title
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    u used the power rule first

    • 2 years ago
  29. swissgirl Group Title
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    I used the chain rule by first bringing down the exponent in front of the bracket and then subtracting one from the exponent as you can see and then I derived what was in the bracket. The derivative of x^2+2 is 2x

    • 2 years ago
  30. swissgirl Group Title
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    Well as we see there are 2 things that need to be derived. The outer exponent and the inner exponent. So the chain rule states that you first derive the outer part and then you derive what is in the bracket

    • 2 years ago
  31. swissgirl Group Title
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    Do you follow @sandy524

    • 2 years ago
  32. swissgirl Group Title
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    alrighty ill just give you the last two lines that i messed up \( \large {2x \over 2\sqrt{x^2+2}}\) \( \large {x \over \sqrt{x^2+2}}\)

    • 2 years ago
  33. sandy524 Group Title
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    i dont get what this means

    • 2 years ago
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