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sandy524

  • 2 years ago

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

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  1. sandy524
    • 2 years ago
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    \[f(x)=\sqrt{x^2+2}\]

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

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

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

  5. sandy524
    • 2 years ago
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    huh

  6. sandy524
    • 2 years ago
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    #4

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

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

  9. zepdrix
    • 2 years ago
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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

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

  11. zepdrix
    • 2 years ago
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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 ^^

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

  13. sandy524
    • 2 years ago
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    lol

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

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

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

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

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

  19. swissgirl
    • 2 years ago
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    uuugghhhhhhhhhh

  20. zepdrix
    • 2 years ago
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    XD

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

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

  23. sandy524
    • 2 years ago
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    :(

  24. sandy524
    • 2 years ago
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    ok thanks

  25. sandy524
    • 2 years ago
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    you guys rock

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

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

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

  29. swissgirl
    • 2 years ago
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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

  30. swissgirl
    • 2 years ago
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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

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

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

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

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